\(L.S=\int{\frac{1}{a^2+x^2}dx}\)
\(=\int{\frac{1}{a^2+a^2\tan^2{\theta}}\times{a\sec^2{\theta}}d\theta}\)
\(=\int{\frac{1}{a^2(1+\tan^2{\theta})}\times{a\sec^2{\theta}}d\theta}\)
\(=\int{\frac{1}{a\sec^2{\theta}}\times{\sec^2{\theta}}d\theta}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{a}d\theta}\)
\(=\frac{1}{a}\int{d\theta}\)
\(=\frac{1}{a}\theta+c\) ➜Note \(\because \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+c\) ➜Note \(\because \theta=\tan^{-1}\left(\frac{x}{a}\right)\)
\(=R.S\)

\(L.S=\int{\frac{1}{\sqrt{a^2-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{a^2-a^2\sin^2{\theta}}}.a\cos{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2(1-\sin^2{\theta})}}.a\cos{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2\cos^2{\theta}}}.a\cos{\theta}d\theta}\) ➜Note \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\frac{1}{a\cos{\theta}}.a\cos{\theta}d\theta}\)
\(=\int{d\theta}\)
\(=\theta+c\) ➜Note \(\because \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}\left(\frac{x}{a}\right)+c\) ➜Note \(\because \theta=\sin^{-1}\left(\frac{x}{a}\right)\)
\(=R.S\)

\(L.S=\int{\frac{1}{x\sqrt{x^2-a^2}}dx}\)
\(=\int{\frac{1}{a\sec{\theta}\sqrt{a^2\sec^2{\theta}-a^2}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2(\sec^2{\theta}-1)}}.\tan{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2\tan^2{\theta}}}.\tan{\theta}d\theta}\) ➜Note \(\because \sec^2{A}-1=\tan^2{A}\)
\(=\int{\frac{1}{a\tan{\theta}}.\tan{\theta}d\theta}\)
\(=\int{\frac{1}{a}.d\theta}\)
\(=\frac{1}{a}\int{d\theta}\)
\(=\frac{1}{a}\theta+c\) ➜Note \(\because \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)+c\) ➜Note \(\because \theta=\sec^{-1}\left(\frac{x}{a}\right)\)
\(=R.S\)

\(L.S=\int{\frac{1}{a^2-x^2}dx}\)
\(=\int{\frac{1}{(a+x)(a-x)}dx}\)
\(=\int{\frac{2a}{2a(a+x)(a-x)}dx}\)
\(=\int{\frac{(a-x)+(a+x)}{2a(a+x)(a-x)}dx}\)
\(=\frac{1}{2a}\int{\frac{(a-x)+(a+x)}{(a+x)(a-x)}dx}\)
\(=\frac{1}{2a}\int{\left\{\frac{(a-x)}{(a+x)(a-x)}+\frac{(a+x)}{(a+x)(a-x)}\right\}dx}\)
\(=\frac{1}{2a}\int{\left\{\frac{1}{a+x}+\frac{1}{a-x}\right\}dx}\)
\(=\frac{1}{2a}\left\{\int{\frac{1}{a+x}dx}+\int{\frac{1}{a-x}dx}\right\}\)
\(=\frac{1}{2a}\left\{\frac{1}{1}\ln{|a+x|}+\frac{1}{-1}\ln{|a-x|}\right\}+c\) ➜Note \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2a}\left\{\ln{|a+x|}-\ln{|a-x|}\right\}+c\)
\(=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}+c\) ➜Note \(\because \ln{|M|}-\ln{|N|}=\ln{\left|\frac{M}{N}\right|}\)
\(=R.S\)

\(L.S=\int{\frac{1}{x^2-a^2}dx}\)
\(=\int{\frac{1}{(x+a)(x-a)}dx}\)
\(=\int{\frac{2a}{2a(x+a)(x-a)}dx}\)
\(=\int{\frac{(x+a)-(x-a)}{2a(x+a)(x-a)}dx}\)
\(=\frac{1}{2a}\int{\frac{(x+a)-(x-a)}{(x+a)(x-a)}dx}\)
\(=\frac{1}{2a}\int{\left\{\frac{(x+a)}{(x+a)(x-a)}-\frac{(x-a)}{(x+a)(x-a)}\right\}dx}\)
\(=\frac{1}{2a}\int{\left\{\frac{1}{x-a}-\frac{1}{x+a}\right\}dx}\)
\(=\frac{1}{2a}\left\{\int{\frac{1}{x-a}dx}+\int{\frac{1}{x+a}dx}\right\}\)
\(=\frac{1}{2a}\left\{\frac{1}{1}\ln{|x-a|}-\frac{1}{1}\ln{|x+a|}\right\}+c\) ➜Note \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2a}\left\{\ln{|x-a|}-\ln{|x+a|}\right\}+c\)
\(=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}+c\) ➜Note \(\because \ln{|M|}-\ln{|N|}=\ln{\left|\frac{M}{N}\right|}\)
\(=R.S\)

\(L.S=\int{\frac{1}{\sqrt{a^2+x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{a^2+a^2\tan^2{\theta}}}.a\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2(1+\tan^2{\theta})}}.a\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2\sec^2{\theta}}}.a\sec^2{\theta}d\theta}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{a\sec{\theta}}.a\sec^2{\theta}d\theta}\)
\(=\int{\sec{\theta}d\theta}\)
\(=\ln{|\sec{\theta}+\tan{\theta}|}+c_{1}\) ➜Note \(\because \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sqrt{1+\tan^2{\theta}}+\tan{\theta}|}+c_{1}\) ➜Note \(\because \sec^2{A}=1+\tan^2{A} \Rightarrow \sec{A}=\sqrt{1+\tan^2{A}}\)
\(=\ln{\left|\sqrt{1+\left(\frac{x}{a}\right)^2}+\frac{x}{a}\right|}+c_{1}\) ➜Note \(\because \tan{\theta}=\frac{x}{a}\)
\(=\ln{\left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|}+c_{1}\)
\(=\ln{\left|\sqrt{\frac{a^2+x^2}{a^2}}+\frac{x}{a}\right|}+c_{1}\)
\(=\ln{\left|\frac{\sqrt{a^2+x^2}}{a}+\frac{x}{a}\right|}+c_{1}\)
\(=\ln{\left|\frac{\sqrt{a^2+x^2}+x}{a}\right|}+c_{1}\)
\(=\ln{|\sqrt{a^2+x^2}+x|}-\ln{a}+c_{1}\) ➜Note \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\ln{|\sqrt{a^2+x^2}+x|}+c_{1}-\ln{a}\)
\(=\ln{|\sqrt{a^2+x^2}+x|}+c\) ➜Note \(\because c_{1}-\ln{a}=c\)
\(=R.S\)

\(L.S=\int{\frac{1}{\sqrt{x^2-a^2}}dx}\)
\(=\int{\frac{1}{\sqrt{a^2\sec^2{\theta}-a^2}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2(\sec^2{\theta}-1)}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\frac{1}{\sqrt{a^2\tan^2{\theta}}}.a\sec{\theta}\tan{\theta}d\theta}\) ➜Note \(\because \sec^2{A}-1=\tan^2{A}\)
\(=\int{\frac{1}{a\tan{\theta}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\sec{\theta}d\theta}\)
\(=\ln{|\sec{\theta}+\tan{\theta}|}+c_{1}\) ➜Note \(\because \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sec{\theta}+\sqrt{\sec^2{\theta}-1}|}+c_{1}\) ➜Note \(\because \tan^2{A}=\sec^2{A}-1 \Rightarrow \tan{A}=\sqrt{\sec^2{A}-1}\)
\(=\ln{\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}\right|}+c_{1}\) ➜Note \(\because \sec{\theta}=\frac{x}{a}\)
\(=\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|}+c_{1}\)
\(=\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2-a^2}{a^2}}\right|}+c_{1}\)
\(=\ln{\left|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}\right|}+c_{1}\)
\(=\ln{\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|}+c_{1}\)
\(=\ln{|x+\sqrt{x^2-a^2}|}-\ln{a}+c_{1}\) ➜Note \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\ln{|\sqrt{x^2-a^2}+x|}+c_{1}-\ln{a}\)
\(=\ln{|\sqrt{x^2-a^2}+x|}+c\) ➜Note \(\because c_{1}-\ln{a}=c\)
\(=R.S\)

\(L.S=\int{\sqrt{a^2-x^2}dx}\)
\(=\int{\sqrt{a^2-a^2\sin^2{\theta}}.a\cos{\theta}d\theta}\)
\(=\int{\sqrt{a^2(1-\sin^2{\theta})}.a\cos{\theta}d\theta}\)
\(=\int{\sqrt{a^2\cos^2{\theta}}.a\cos{\theta}d\theta}\) ➜Note \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{a\cos{\theta}.a\cos{\theta}d\theta}\)
\(=a^2\int{\cos^2{\theta}d\theta}\)
\(=\frac{a^2}{2}\int{2\cos^2{\theta}d\theta}\)
\(=\frac{a^2}{2}\int{(1+\cos{2\theta})d\theta}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{a^2}{2}\left\{\theta+\frac{\sin{2\theta}}{2}\right\}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{\sin{ax}}{a} \) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a^2}{2}\left\{\theta+\frac{2\sin{\theta}\cos{\theta}}{2}\right\}+c\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=\frac{a^2}{2}\left\{\theta+\sin{\theta}\cos{\theta}\right\}+c\)
\(=\frac{a^2}{2}\left\{\theta+\sin{\theta}\sqrt{1-\sin^2{\theta}}\right\}+c\) ➜Note \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
\(=\frac{a^2}{2}\left\{\sin^{-1}\left(\frac{x}{a}\right)+\frac{x}{a}\sqrt{1-\left(\frac{x}{a}\right)^2}\right\}+c\) ➜Note \(\because \theta=\sin^{-1}\left(\frac{x}{a}\right), \sin{\theta}=\frac{x}{a}\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}\sqrt{\frac{a^2-x^2}{a^2}}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}.\frac{\sqrt{a^2-x^2}}{a}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x\sqrt{a^2-x^2}}{a^2}\right\}+c\)
\(=\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x\sqrt{a^2-x^2}}{2}+c\)
\(=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}+c\)
\(=R.S\)

\(\int{\frac{1}{x^2+8x+25}dx}\)
\(=\int{\frac{1}{x^2+2.x.4+4^2-4^2+25}dx}\)
\(=\int{\frac{1}{(x+4)^2-16+25}dx}\)
\(=\int{\frac{1}{(x+4)^2+9}dx}\)
\(=\int{\frac{1}{(t)^2+9}dt}\)
\(=\int{\frac{1}{(t)^2+3^2}dt}\)
\(=\int{\frac{1}{3^2+(t)^2}dt}\)
\(=\frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right)+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan^{-1}\left(\frac{x+4}{3}\right)+c\) ➜Note \(\because t=x+4\)

\(\int{\frac{1}{(2x+3)\sqrt{4x+5}}dx}\)
\(=\int{\frac{1}{\left\{2\left(\frac{1}{4}t^2-\frac{5}{4}\right)+3\right\}t}.\frac{1}{2}tdt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{1}{2}t^2-\frac{5}{2}+3}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{1}{2}t^2+\frac{-5+6}{2}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{1}{2}t^2+\frac{1}{2}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{t^2+1}{2}}dt}\)
\(=\frac{1}{2}\int{\frac{2}{t^2+1}dt}\)
\(=\int{\frac{1}{t^2+1}dt}\)
\(=\int{\frac{1}{1+t^2}dt}\)
\(=\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan^{-1}{(\sqrt{4x+5})}+c\) ➜Note \(\because t=\sqrt{4x+5}\)

\(\int{\frac{1}{(x-1)\sqrt{x^2+1}}dx}\)
\(=\int{\frac{1}{\frac{1}{t}\sqrt{\left(\frac{1}{t}+1\right)^2+1}}\times{-\frac{1}{t^2}dt}}\)
\(=-\int{\frac{1}{\sqrt{\left(\frac{1}{t}+1\right)^2+1}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\sqrt{\frac{1}{t^2}+2\frac{1}{t}+1+1}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\sqrt{\frac{1}{t^2}+2\frac{1}{t}+2}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\sqrt{\frac{1+2t+2t^2}{t^2}}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\frac{1}{t}\sqrt{1+2t+2t^2}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\sqrt{1+2t+2t^2}}dt}\)
\(=-\int{\frac{1}{\sqrt{2\left(t^2+t+\frac{1}{2}\right)}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{t^2+t+\frac{1}{2}}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{t^2+2.t.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}+\frac{1}{2}}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2+\frac{1}{2}-\frac{1}{4}}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2+\frac{2-1}{4}}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2+\frac{1}{4}}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}}dt}\)
\(=-\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(\frac{1}{2}\right)^2+\left(t+\frac{1}{2}\right)^2}}dt}\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{\left(\frac{1}{2}\right)^2+\left(t+\frac{1}{2}\right)^2}+\left(t+\frac{1}{2}\right)\right|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2+x^2}}dx}=\ln{|\sqrt{a^2+x^2}+x|}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{t^2+t+\frac{1}{4}+\frac{1}{4}}+\left(t+\frac{1}{2}\right)\right|}+c\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{t^2+t+\frac{1+1}{4}}+\left(t+\frac{1}{2}\right)\right|}+c\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{t^2+t+\frac{2}{4}}+\left(t+\frac{1}{2}\right)\right|}+c\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{t^2+t+\frac{1}{2}}+\left(t+\frac{1}{2}\right)\right|}+c\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\sqrt{\left(\frac{1}{x-1}\right)^2+\left(\frac{1}{x-1}\right)+\frac{1}{2}}+\left(\frac{1}{x-1}+\frac{1}{2}\right)\right|}+c\)
\(=-\frac{1}{\sqrt{2}}\ln{\left|\left(\frac{1}{x-1}+\frac{1}{2}\right)+\sqrt{\left(\frac{1}{x-1}\right)^2+\left(\frac{1}{x-1}\right)+\frac{1}{2}}\right|}+c\) ➜Note \(\because x-1=\frac{1}{t}\Rightarrow t=\frac{1}{x-1}\)

\(\int{\frac{1}{(x^2+1)\sqrt{x^2+4}}dx}\)
\(=\int{\frac{1}{(x^2+1)}.\frac{dx}{\sqrt{x^2+4}}}\)
\(=\int{\frac{1}{\frac{t^2+3}{t^2-1}}.\frac{dt}{(1-t^2)}}\)
\(=\int{\frac{t^2-1}{t^2+3}.\frac{dt}{(1-t^2)}}\)
\(=-\int{\frac{1-t^2}{t^2+3}.\frac{dt}{(1-t^2)}}\)
\(=-\int{\frac{1}{t^2+3}dt}\)
\(=-\int{\frac{1}{3+t^2}dt}\)
\(=-\int{\frac{1}{(\sqrt{3})^2+t^2}dt}\)
\(=-\frac{1}{\sqrt{3}}\tan^{-1}{\left(\frac{t}{\sqrt{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{\sqrt{3}}\tan^{-1}{\left(\frac{\frac{\sqrt{x^2+4}}{x}}{\sqrt{3}}\right)}+c\) ➜Note \(\because t=\frac{\sqrt{x^2+4}}{x}\)
\(=-\frac{1}{\sqrt{3}}\tan^{-1}{\left(\frac{\sqrt{x^2+4}}{\sqrt{3}x}\right)}+c\)

\(\int{\frac{1}{\sqrt{\sin^3{x}}\sqrt{\cos^5{x}}}dx}\)
\(=\int{\frac{\sec^4{x}}{\sqrt{\sin^3{x}}\sqrt{\cos^5{x}}}\sec^4{x}dx}\)
\(=\int{\frac{\sec^4{x}}{\sqrt{\sin^3{x}}\sqrt{\cos^5{x}}}\sqrt{\sec^8{x}}dx}\)
\(=\int{\frac{\sec^4{x}}{\sqrt{\sin^3{x}}\sqrt{\sec^3{x}}\sqrt{\cos^5{x}}}\sqrt{\sec^5{x}}dx}\)
\(=\int{\frac{\sec^2{x}\sec^2{x}}{\sqrt{\frac{\sin^3{x}}{\cos^3{x}}}}dx}\)
\(=\int{\frac{(1+tan^2{x})\sec^2{x}}{\sqrt{\tan^3{x}}}dx}\)
\(=\int{\frac{(1+tan^2{x})}{\sqrt{\tan^3{x}}}\sec^2{x}dx}\)
\(=\int{\frac{1+t^2}{\sqrt{t^3}}dt}\)
\(=\int{\frac{1+t^2}{t^{\frac{3}{2}}}dt}\)
\(=\int{\left(\frac{1}{t^{\frac{3}{2}}}+\frac{t^2}{t^{\frac{3}{2}}}\right)dt}\)
\(=\int{\left(t^{-\frac{3}{2}}+t^{2-\frac{3}{2}}\right)dt}\)
\(=\int{\left(t^{-\frac{3}{2}}+t^{\frac{4-3}{2}}\right)dt}\)
\(=\int{\left(t^{-\frac{3}{2}}+t^{\frac{1}{2}}\right)dt}\)
\(=\int{t^{-\frac{3}{2}}dt}+\int{t^{\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-3+2}{2}}}{\frac{-3+2}{2}}+\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{t^{\frac{1}{2}}}+\frac{2}{3}\sqrt{t^3}+c\)
\(=-\frac{2}{\sqrt{t}}+\frac{2}{3}\sqrt{t^3}+c\)
\(=-\frac{2}{\sqrt{\tan{x}}}+\frac{2}{3}\sqrt{\tan^3{x}}+c\) ➜Note \(\because t=\tan{x}\)

\(\int{\frac{\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx}\)
\(=\int{\frac{\sin{x}\cos{x}\sec^4{x}}{\sec^4{x}(\sin^4{x}+\cos^4{x})}dx}\)
\(=\int{\frac{\sin{x}\cos{x}\sec^2{x}\sec^2{x}}{\sec^4{x}(\sin^4{x}+\cos^4{x})}dx}\)
\(=\int{\frac{\sin{x}\cos{x}\frac{1}{\cos^2{x}}\sec^2{x}}{\frac{1}{\cos^4{x}}(\sin^4{x}+\cos^4{x})}dx}\)
\(=\int{\frac{\frac{\sin{x}}{\cos{x}}\sec^2{x}}{\frac{\sin^4{x}}{\cos^4{x}}+1}dx}\)
\(=\int{\frac{\tan{x}\sec^2{x}}{\tan^4{x}+1}dx}\)
\(=\int{\frac{\tan{x}}{\tan^4{x}+1}\sec^2{x}dx}\)
\(=\int{\frac{t}{t^4+1}dt}\)
\(=\int{\frac{1}{(t^2)^2+1}tdt}\)
\(=\int{\frac{1}{z^2+1}.\frac{1}{2}dz}\)
\(=\frac{1}{2}\int{\frac{1}{1+z^2}dz}\)
\(=\frac{1}{2}\tan^{-1}{z}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\tan^{-1}{t^2}+c\) ➜Note \(\because z=t^2\)
\(=\frac{1}{2}\tan^{-1}{\tan^2{x}}+c\) ➜Note \(\because t=\tan{x}\)

\(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\)
\(=\int{\frac{dx}{(x^{\frac{1}{4}})^2-x^{\frac{1}{4}}}}\)
\(=\int{\frac{4t^3dt}{t^2-t}}\)
\(=4\int{\frac{t^3dt}{t(t-1)}}\)
\(=4\int{\frac{t^2}{t-1}dt}\)
\(=4\int{\frac{t^2-1+1}{t-1}dt}\)
\(=4\int{\frac{(t+1)(t-1)+1}{t-1}dt}\)
\(=4\int{\left\{\frac{(t+1)(t-1)}{t-1}+\frac{1}{t-1}\right\}dt}\)
\(=4\int{\left\{t+1+\frac{1}{t-1}\right\}dt}\)
\(=4\int{tdt}+4\int{dt}+4\int{\frac{1}{t-1}dt}\)
\(=4\frac{t^2}{2}+4t+4\ln{|t-1|}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2t^2+4t+4\ln{|t-1|}+c\)
\(=2(x^{\frac{1}{4}})^2+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\) ➜Note \(\because t=x^{\frac{1}{4}}\)
\(=2x^{\frac{1}{2}}+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\)
\(=2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)

\(\int{\sqrt{\frac{1-x}{1+x}}dx}\)
\(=\int{\frac{\sqrt{1-x}}{\sqrt{1+x}}dx}\)
\(=\int{\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1+x}.\sqrt{1-x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{1-x}\) গুণ করে।
\(=\int{\frac{1-x}{\sqrt{(1+x)(1-x)}}dx}\)
\(=\int{\frac{1-x}{\sqrt{1-x^2}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}-\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}-\int{\frac{1}{t}\times{-tdt}}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}+\int{dt}\)
\(=\sin^{-1}{x}+t+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{x}+\sqrt{1-x^2}+c\) ➜Note \(\because t=\sqrt{1-x^2}\)

\(\int{\frac{dx}{\sqrt{(x-2)(x-3)}}}\)
\(=\int{\frac{2}{t}dt}\)
\(=2\int{\frac{1}{t}dt}\)
\(=2\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{|\sqrt{x-2}+\sqrt{x-3}|}+c\) ➜Note \(\because t=\sqrt{25-x^2}\)

\(\int{\frac{dx}{\sqrt{(x-2)(3-x)}}}\)
\(=\int{\frac{1}{\sqrt{(x-2)}\sqrt{(3-x)}}dx}\)
\(=\int{\frac{1}{t\sqrt{3-(t^2+2)}}2tdt}\)
\(=2\int{\frac{1}{\sqrt{3-t^2-2}}dt}\)
\(=2\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=2\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sin^{-1}{(\sqrt{x-2})}+c\) ➜Note \(\because t=\sqrt{x-2}\)

\(\int{\frac{dx}{3+2\sin{x}}}\)
\(=\int{\frac{1}{3+2\sin{x}}dx}\)
\(=\int{\frac{1}{3+2\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{3(1+\tan^2{\frac{x}{2}})+4\tan{\frac{x}{2}}}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{3+3\tan^2{\frac{x}{2}}+4\tan{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{3\tan^2{\frac{x}{2}}+4\tan{\frac{x}{2}}+3}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{3t^2+4t+3}2dt}\)
\(=2\int{\frac{1}{3\left(t^2+\frac{4}{3}t+1\right)}dt}\)
\(=\frac{2}{3}\int{\frac{1}{t^2+\frac{4}{3}t+1}dt}\)
\(=\frac{2}{3}\int{\frac{1}{t^2+2.t.\frac{2}{3}+\left(\frac{2}{3}\right)^2-\frac{4}{9}+1}dt}\)
\(=\frac{2}{3}\int{\frac{1}{\left(t+\frac{2}{3}\right)^2+1-\frac{4}{9}}dt}\)
\(=\frac{2}{3}\int{\frac{1}{\left(t+\frac{2}{3}\right)^2+\frac{9-4}{9}}dt}\)
\(=\frac{2}{3}\int{\frac{1}{\left(t+\frac{2}{3}\right)^2+\frac{5}{9}}dt}\)
\(=\frac{2}{3}\int{\frac{1}{\frac{5}{9}+\left(t+\frac{2}{3}\right)^2}dt}\)
\(=\frac{2}{3}\int{\frac{1}{\left(\frac{\sqrt{5}}{3}\right)^2+\left(t+\frac{2}{3}\right)^2}dt}\)
\(=\frac{2}{3}.\frac{1}{\frac{\sqrt{5}}{3}}\tan^{-1}{\frac{t+\frac{2}{3}}{\frac{\sqrt{5}}{3}}}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{3}.\frac{3}{\sqrt{5}}\tan^{-1}{\frac{3t+2}{\sqrt{5}}}+c\) ➜Note লব ও হরের সহিত \(3\) গুণ করে।
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\frac{3\tan{\frac{x}{2}}+2}{\sqrt{5}}}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left\{\frac{1}{\sqrt{5}}(3\tan{\frac{x}{2}}+2)\right\}}+c\)

\(\int{\frac{dx}{3+2\cos{x}}}\)
\(=\int{\frac{1}{3+2\cos{x}}dx}\)
\(=\int{\frac{1}{3+2\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{3(1+\tan^2{\frac{x}{2}})+2(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{3+3\tan^2{\frac{x}{2}}+2-2\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{5+\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{5+t^2}2dt}\)
\(=2\int{\frac{1}{\left\{5+t^2\right\}}dt}\)
\(=2\int{\frac{1}{(\sqrt{5})^2+t^2}dt}\)
\(=2.\frac{1}{\sqrt{5}}\tan^{-1}{\left(\frac{t}{\sqrt{5}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left(\frac{t}{\sqrt{5}}\right)}+c\)
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left(\frac{\tan{\frac{x}{2}}}{\sqrt{5}}\right)}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)

\(\int{\frac{dx}{\sin{x}-\cos{x}+1}}\)
\(=\int{\frac{dx}{1+\sin{x}-\cos{x}}}\)
\(=\int{\frac{1}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\frac{1}{1+\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)-\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-1+\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{2t^2+2t}2dt}\)
\(=\int{\frac{1}{2(t^2+t)}2dt}\)
\(=\int{\frac{1}{t^2+t}dt}\)
\(=\int{\frac{1}{t^2+2.t.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}dt}\)
\(=\frac{1}{2.\frac{1}{2}}\ln{\left|\frac{t+\frac{1}{2}-\frac{1}{2}}{t+\frac{1}{2}+\frac{1}{2}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{1}\ln{\left|\frac{t}{t+\frac{1+1}{2}}\right|}+c\)
\(=\ln{\left|\frac{t}{t+\frac{2}{2}}\right|}+c\)
\(=\ln{\left|\frac{t}{t+1}\right|}+c\)
\(=\ln{\left|\frac{\tan{\frac{x}{2}}}{\tan{\frac{x}{2}}+1}\right|}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)
\(=\ln{\left|\frac{\tan{\frac{x}{2}}}{1+\tan{\frac{x}{2}}}\right|}+c\)

\(\int{\frac{dx}{a\cos{x}+b\sin{x}}}\)
\(=\int{\frac{1}{r\sin{\alpha}\cos{x}+r\cos{\alpha}\sin{x}}dx}\) ➜Note \(\because a=r\sin{\alpha}, b=r\cos{\alpha}\)
\(=\int{\frac{1}{r(\sin{\alpha}\cos{x}+\cos{\alpha}\sin{x})}dx}\)
\(=\int{\frac{1}{r\sin{(\alpha+x)}}dx}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(=\frac{1}{r}\int{\frac{1}{\sin{(x+\alpha)}}dx}\)
\(=\frac{1}{r}\int{cosec \ {(x+\alpha)}dx}\) ➜Note \(\because \frac{1}{\sin{A}}=cosec \ {A}\)
\(=\frac{1}{r}\ln{\left|\tan{\left(\frac{x+\alpha}{2}\right)}\right|}+c\) ➜Note \(\because \int{cosec \ {x}dx}=\ln{\left|\tan{\left(\frac{x}{2}\right)}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{a^2+b^2}}\ln{\left|\tan{\frac{x+\tan^{-1}{\left(\frac{a}{b}\right)}}{2}}\right|}+c\) ➜Note \(\because \alpha=\tan^{-1}{\left(\frac{a}{b}\right)}, r=\sqrt{a^2+b^2}\)
\(=\frac{1}{\sqrt{a^2+b^2}}\ln{\left|\tan{\frac{1}{2}\left\{x+\tan^{-1}{\left(\frac{a}{b}\right)}\right\}}\right|}+c\)

ধরি,
\(2\sin{x}+3\cos{x}=L(7\sin{x}-2\cos{x})\)\(+M\left\{\frac{d}{dx}(7\sin{x}-2\cos{x})\right\}\)
\(\Rightarrow 2\sin{x}+3\cos{x}=L(7\sin{x}-2\cos{x})+M(7\cos{x}+2\sin{x}) ....(1)\)
\(=7L\sin{x}-2L\cos{x}+7M\cos{x}+2M\sin{x}\)
\(=(7L+2M)\sin{x}+(7M-2L)\cos{x}\)
\(\therefore 2\sin{x}+3\cos{x}=(7L+2M)\sin{x}+(7M-2L)\cos{x}\)
\(\Rightarrow (7L+2M)\sin{x}+(7M-2L)\cos{x}=2\sin{x}+3\cos{x} ....(2)\)
\(\Rightarrow 7L+2M=2 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 7M-2L=3.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\((3)\Rightarrow 7L+2M-2=0\)
\((4)\Rightarrow -2L+7M-3=0\)
\(\Rightarrow \frac{L}{-6+14}=\frac{M}{4+21}=\frac{1}{49+4}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{8}=\frac{M}{25}=\frac{1}{53}\)
\(\Rightarrow \frac{L}{8}=\frac{1}{53}; \frac{M}{25}=\frac{1}{53}\)
\(\therefore L=\frac{8}{53}; M=\frac{25}{53}\)
\(L\) ও \(M\) এর মান \((1)\) এ বসিয়ে,
\(2\sin{x}+3\cos{x}=\frac{8}{53}(7\sin{x}-2\cos{x})+\frac{25}{53}(7\cos{x}+2\sin{x})\)
\(\int{\frac{2\sin{x}+3\cos{x}}{7\sin{x}-2\cos{x}}dx}\)
\(=\int{\frac{\frac{8}{53}(7\sin{x}-2\cos{x})+\frac{25}{53}(7\cos{x}+2\sin{x})}{7\sin{x}-2\cos{x}}dx}\)
\(=\int{\left\{\frac{8}{53}+\frac{25}{53}\frac{(7\cos{x}+2\sin{x})}{7\sin{x}-2\cos{x}}\right\}dx}\)
\(=\frac{8}{53}\int{dx}+\frac{25}{53}\int{\frac{(7\cos{x}+2\sin{x})}{7\sin{x}-2\cos{x}}dx}\)
\(=\frac{8}{53}\int{dx}+\frac{25}{53}\int{\frac{1}{7\sin{x}-2\cos{x}}(7\cos{x}+2\sin{x})dx}\)
\(=\frac{8}{53}\int{dx}+\frac{25}{53}\int{\frac{1}{t}dt}\)
\(=\frac{8}{53}x+\frac{25}{53}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{8x}{53}+\frac{25}{53}\ln{|7\sin{x}-2\cos{x}|}+c\) ➜Note \(\because t=7\sin{x}-2\cos{x}\)

ধরি,
\(1-\sin{x}+\cos{x}=L(1+\sin{x}-\cos{x})\)\(+M\left\{\frac{d}{dx}(1+\sin{x}-\cos{x})\right\}+N\)
\(\Rightarrow 1-\sin{x}+\cos{x}=L(1+\sin{x}-\cos{x})+M(\cos{x}+\sin{x})+N ....(1)\)
\(=L+L\sin{x}-L\cos{x}+M\cos{x}+M\sin{x}+N\)
\(=(L+M)\sin{x}+(M-L)\cos{x}+L+N\)
\(\therefore 1-\sin{x}+\cos{x}=(L+M)\sin{x}+(M-L)\cos{x}+L+N\)
\(\Rightarrow (L+M)\sin{x}+(M-L)\cos{x}+L+N=-\sin{x}+\cos{x}+1 ........(2)\)
\(\Rightarrow L+M=-1 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow M-L=1.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow L+N=1.....(5)\) ➜Note \((2)\) এর উভয় পার্শ হতে ধ্রুবক রাশির সমতা নিয়ে ।
\((3)+(4)\) এর সাহায্যে
\(L+M+M-L=-1+1\)
\(\Rightarrow 2M=0\)
\(\therefore M=0\)
\((3)\) হতে,
\(L+0=-1\)
\(\Rightarrow L=-1\)
\((5)\) হতে,
\(-1+N=1\)
\(\Rightarrow N=1+1\)
\(\therefore N=2\)
\(L\), \(M\) ও \(N\) এর মান \((1)\) এ বসিয়ে,
\(1-\sin{x}+\cos{x}=-(1+\sin{x}-\cos{x})+0.(\cos{x}+\sin{x})+2\)
\(\Rightarrow 1-\sin{x}+\cos{x}=-(1+\sin{x}-\cos{x})+2\)
\(\int{\frac{1-\sin{x}+\cos{x}}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\frac{-(1+\sin{x}-\cos{x})+2}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\left\{\frac{-(1+\sin{x}-\cos{x})}{1+\sin{x}-\cos{x}}+\frac{2}{1+\sin{x}-\cos{x}}\right\}dx}\)
\(=-\int{dx}+2\int{\frac{1}{1+\sin{x}-\cos{x}}dx}\)
\(=-\int{dx}+2\int{\frac{1}{1+\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)-\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=-\int{dx}+2\int{\frac{1+\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=-\int{dx}+2\int{\frac{\sec^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-1+\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=-\int{dx}+2\int{\frac{1}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=-\int{dx}+2\int{\frac{1}{2t^2+2t}2dt}\)
\(=-\int{dx}+2\int{\frac{1}{2(t^2+t)}2dt}\)
\(=-\int{dx}+2\int{\frac{1}{t^2+t}dt}\)
\(=-\int{dx}+2\int{\frac{1}{t^2+2.t.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=-\int{dx}+2\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=-\int{dx}+2\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}dt}\)
\(=-x+2.\frac{1}{2.\frac{1}{2}}\ln{\left|\frac{t+\frac{1}{2}-\frac{1}{2}}{t+\frac{1}{2}+\frac{1}{2}}\right|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-x+2.\frac{1}{1}\ln{\left|\frac{t}{t+\frac{1+1}{2}}\right|}+c\)
\(=-x+2\ln{\left|\frac{t}{t+\frac{2}{2}}\right|}+c\)
\(=-x+2\ln{\left|\frac{t}{t+1}\right|}+c\)
\(=-x+2\ln{\left|\frac{\tan{\frac{x}{2}}}{\tan{\frac{x}{2}}+1}\right|}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)
\(=-x+2\ln{\left|\frac{\tan{\frac{x}{2}}}{1+\tan{\frac{x}{2}}}\right|}+c\)

\(\int{\frac{dx}{(x-b)^3(x-a)^2}}\)
\(=\int{\frac{\frac{x-a}{1-t}dt}{t^3(x-a)^3(x-a)^2}}\)
\(=\int{\frac{(x-a)dt}{(1-t)t^3(x-a)^5}}\)
\(=\int{\frac{dt}{(1-t)t^3(x-a)^4}}\)
\(=\int{\frac{dt}{(1-t)t^3\left(-\frac{a-b}{1-t}\right)^4}}\)
\(=\int{\frac{dt}{(1-t)t^3\times{\frac{(a-b)^4}{(1-t)^4}}}}\)
\(=\int{\frac{(1-t)^4dt}{(1-t)t^3(a-b)^4}}\)
\(=\frac{1}{(a-b)^4}\int{\frac{(1-t)^3dt}{t^3}}\)
\(=\frac{1}{(a-b)^4}\int{\frac{1-3t+3t^2-t^3}{t^3}dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(\frac{1}{t^3}-3\frac{t}{t^3}+3\frac{t^2}{t^3}-\frac{t^3}{t^3}\right)dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(\frac{1}{t^3}-3\frac{1}{t^2}+3\frac{1}{t}-1\right)dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(t^{-3}-3t^{-2}+3\frac{1}{t}-1\right)dt}\)
\(=\frac{1}{(a-b)^4}\left[\int{t^{-3}dt}-3\int{t^{-2}dt}+3\int{\frac{1}{t}dt}-\int{dt}\right]\)
\(=\frac{1}{(a-b)^4}\left[\frac{t^{-3+1}}{-3+1}-3\frac{t^{-2+1}}{-2+1}+3\ln{|t|}-t\right]+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\frac{1}{x}dx}=\ln{|x|}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{(a-b)^4}\left[-\frac{1}{2t^2}+\frac{3}{t}+3\ln{|t|}-t\right]+c\)
\(=\frac{1}{(a-b)^4}\left[-\frac{1}{2\left(\frac{x-b}{x-a}\right)^2}+\frac{3}{\frac{x-b}{x-a}}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{x-b}{x-a}\right]+c\) ➜Note \(\because x-b=t(x-a)\Rightarrow \frac{x-b}{x-a}=t\)
\(=\frac{1}{(a-b)^4}\left[-\frac{(x-a)^2}{2(x-b)^2}+\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{x-b}{x-a}\right]+c\)
\(=\frac{1}{(a-b)^4}\left[\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{(x-a)^2}{2(x-b)^2}-\frac{x-b}{x-a}\right]+c\)

\(\int{\frac{dx}{x^2+25}}\)
\(=\int{\frac{dx}{x^2+5^2}}\)
\(=\int{\frac{dx}{5^2+x^2}}\)
\(=\frac{1}{5}\tan^{-1}{\left(\frac{x}{5}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।

\(\int{\frac{dx}{\sqrt{9-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{4\left(\frac{9}{4}-x^2\right)}}}\)
\(=\int{\frac{dx}{2\sqrt{\left(\frac{9}{4}-x^2\right)}}}\) \(=\frac{1}{2}\int{\frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2-x^2}}}\)
\(=\frac{1}{2}\sin^{-1}{\left(\frac{x}{\frac{3}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sin^{-1}{\left(\frac{2x}{3}\right)}+c\)

\(\int{\frac{1}{9x^2-16}dx}\)
\(=\int{\frac{1}{9\left(x^2-\frac{16}{9}\right)}dx}\)
\(=\frac{1}{9}\int{\frac{1}{x^2-\left(\frac{4}{3}\right)^2}dx}\)
\(=\frac{1}{9}\times{\frac{1}{2.\frac{4}{3}}}\ln{\left|\frac{x-\frac{4}{3}}{x+\frac{4}{3}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{9}\times{\frac{3}{8}}\ln{\left|\frac{\frac{3x-4}{3}}{\frac{3x+4}{3}}\right|}+c\)
\(=\frac{1}{3}\times{\frac{1}{8}}\ln{\left|\frac{3x-4}{3}\times{\frac{3}{3x+4}}\right|}+c\)
\(=\frac{1}{24}\ln{\left|\frac{3x-4}{3x+4}\right|}+c\)

\(\int{\frac{1}{x^2+x+1}dx}\)
\(=\int{\frac{1}{x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}+1}dx}\)
\(=\int{\frac{1}{\left(x+\frac{1}{2}\right)^2+1-\frac{1}{4}}dx}\)
\(=\int{\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{4-1}{4}}dx}\)
\(=\int{\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx}\)
\(=\int{\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2+\left(x+\frac{1}{2}\right)^2}dx}\)
\(=\int{\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2+t^2}dt}\)
\(=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}{\left(\frac{t}{\frac{\sqrt{3}}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\left(\frac{2t}{\sqrt{3}}\right)}+c\)
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\left(\frac{2\left(x+\frac{1}{2}\right)}{\sqrt{3}}\right)}+c\) ➜Note \(\because t=x+\frac{1}{2}\)
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}+c\)

\(\int{\frac{xdx}{2x^4-3x^2-2}}\)
\(=\int{\frac{xdx}{2\left(x^4-\frac{3}{2}x^2-1\right)}}\)
\(=\frac{1}{2}\int{\frac{xdx}{(x^2)^2-2.x^2.\frac{3}{4}+\left(\frac{3}{4}\right)^2-\frac{9}{16}-1}}\)
\(=\frac{1}{2}\int{\frac{xdx}{\left(x^2-\frac{3}{4}\right)^2-\left(\frac{9}{16}+1\right)}}\)
\(=\frac{1}{2}\int{\frac{xdx}{\left(x^2-\frac{3}{4}\right)^2-\left(\frac{9+16}{16}\right)}}\)
\(=\frac{1}{2}\int{\frac{xdx}{\left(x^2-\frac{3}{4}\right)^2-\left(\frac{25}{16}\right)}}\)
\(=\frac{1}{2}\int{\frac{1}{\left(x^2-\frac{3}{4}\right)^2-\left(\frac{5}{4}\right)^2}xdx}\)
\(=\frac{1}{2}\int{\frac{1}{t^2-\left(\frac{5}{4}\right)^2}\frac{1}{2}dt}\)
\(=\frac{1}{4}\int{\frac{1}{t^2-\left(\frac{5}{4}\right)^2}dt}\)
\(=\frac{1}{4}.\frac{1}{2.\frac{5}{4}}\ln{\left|\frac{t-\frac{5}{4}}{t+\frac{5}{4}}\right|}+c^{\prime}\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c^{\prime}\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}.\frac{1}{\frac{5}{2}}\ln{\left|\frac{4t-5}{4t+5}\right|}+c^{\prime}\) ➜Note লব ও হরের সহিত \(4\) গুণ করে।
\(=\frac{1}{4}.\frac{2}{5}\ln{\left|\frac{4t-5}{4t+5}\right|}+c^{\prime}\)
\(=\frac{1}{2}.\frac{1}{5}\ln{\left|\frac{4t-5}{4t+5}\right|}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left|\frac{4\left(x^2-\frac{3}{4}\right)-5}{4\left(x^2-\frac{3}{4}\right)+5}\right|}+c^{\prime}\) ➜Note \(\because t=x^2-\frac{3}{4}\)
\(=\frac{1}{10}\ln{\left|\frac{4x^2-3-5}{4x^2-3+5}\right|}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left|\frac{4x^2-8}{4x^2+2}\right|}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left|\frac{2x^2-4}{2x^2+1}\right|}+c^{\prime}\) ➜Note লব ও হরের সহিত \(2\) ভাগ করে।
\(=\frac{1}{10}\ln{\left|\frac{2(x^2-2)}{2x^2+1}\right|}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left\{\left|\frac{x^2-2}{2x^2+1}\right|\times{2}\right\}}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left|\frac{x^2-2}{2x^2+1}\right|}+\frac{1}{10}\ln{2}+c^{\prime}\)
\(=\frac{1}{10}\ln{\left|\frac{x^2-2}{2x^2+1}\right|}+c\) ➜Note \(\because \frac{1}{10}\ln{2}+c^{\prime}=c\)

\(\int{\frac{e^xdx}{5-4e^x-e^{2x}}}\)
\(=\int{\frac{}{5-4e^x-(e^{x})^2}e^xdx}\)
\(=\int{\frac{1}{5-4t-t^2}dt}\)
\(=\int{\frac{1}{9-4-4t-t^2}dt}\)
\(=\int{\frac{1}{9-(t^2+4t+4)}dt}\)
\(=\int{\frac{1}{9-(t^2+2.t.2+2^2)}dt}\)
\(=\int{\frac{1}{3^2-(t+2)^2}dt}\)
\(=\int{\frac{1}{3^2-z^2}dz}\)
\(=\frac{1}{2.3}\ln{\left|\frac{3+z}{3-z}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{6}\ln{\left|\frac{3+t+2}{3-t-2}\right|}+c\) ➜Note \(\because z=t+2\)
\(=\frac{1}{6}\ln{\left|\frac{5+t}{1-t}\right|}+c\)
\(=\frac{1}{6}\ln{\left|\frac{5+e^x}{1-e^x}\right|}+c\) ➜Note \(\because t=e^x\)

\(\int{\frac{(3x-5)dx}{x^2-2x+10}}\)
\(=\int{\frac{\frac{3}{2}(2x-2)-2}{x^2-2x+10}dx}\)
\(=\int{\left\{\frac{\frac{3}{2}(2x-2)}{x^2-2x+10}-\frac{2}{x^2-2x+10}\right\}dx}\)
\(=\int{\frac{\frac{3}{2}(2x-2)}{x^2-2x+10}dx}-\int{\frac{2}{x^2-2x+10}dx}\)
\(=\frac{3}{2}\int{\frac{(2x-2)}{x^2-2x+10}dx}-2\int{\frac{1}{x^2-2x+10}dx}\)
\(=\frac{3}{2}\int{\frac{(2x-2)}{x^2-2x+10}dx}-2\int{\frac{1}{x^2-2x+1+9}dx}\)
\(=\frac{3}{2}\int{\frac{(2x-2)}{x^2-2x+10}dx}-2\int{\frac{1}{(x-1)^2+9}dx}\)
\(=\frac{3}{2}\int{\frac{(2x-2)}{x^2-2x+10}dx}-2\int{\frac{1}{(x-1)^2+3^2}dx}\)
\(=\frac{3}{2}\int{\frac{(2x-2)}{x^2-2x+10}dx}-2\int{\frac{1}{t^2+3^2}dt}\)
\(=\frac{3}{2}\int{\frac{d(x^2-2x+10)}{x^2-2x+10}}-2\int{\frac{1}{3^2+t^2}dt}\)
\(=\frac{3}{2}\ln{\left|x^2-2x+10\right|}-2\frac{1}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\) ➜Note \(\because \int{\frac{dx}{x}}=\ln{x}, \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{2}\ln{\left|x^2-2x+10\right|}-\frac{2}{3}\tan^{-1}{\left(\frac{x-1}{3}\right)}+c\)

\(\int{\frac{dx}{\sqrt{2x^2+3x+4}}}\)
\(=\int{\frac{dx}{\sqrt{2\left(x^2+\frac{3}{2}x+2\right)}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{x^2+2.x.\frac{3}{4}+\left(\frac{3}{4}\right)^2-\frac{9}{16}+2}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\left(x+\frac{3}{4}\right)^2+2-\frac{9}{16}}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{32-9}{16}}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{23}{16}}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dt}{\sqrt{t^2+\left(\frac{\sqrt{23}}{4}\right)^2}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dt}{\sqrt{\left(\frac{\sqrt{23}}{4}\right)^2}+t^2}}\)
\(=\frac{1}{\sqrt{2}}\ln{\left|\sqrt{t^2+\left(\frac{\sqrt{23}}{4}\right)^2}+t\right|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2+x^2}}dx}=\ln{\left|\sqrt{a^2+x^2}+x\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\ln{\left|\sqrt{\left(x+\frac{3}{4}\right)^2+\frac{23}{16}}+x+\frac{3}{4}\right|}+c\) ➜Note \(\because t=x+\frac{3}{4}\)
\(=\frac{1}{\sqrt{2}}\ln{\left|x+\frac{3}{4}+\sqrt{x^2+2.x.\frac{3}{4}+\frac{9}{16}+\frac{23}{16}}\right|}+c\)
\(=\frac{1}{\sqrt{2}}\ln{\left|x+\frac{3}{4}+\sqrt{x^2+\frac{3}{2}x+\frac{9+23}{16}}\right|}+c\)
\(=\frac{1}{\sqrt{2}}\ln{\left|x+\frac{3}{4}+\sqrt{x^2+\frac{3}{2}x+\frac{32}{16}}\right|}+c\)
\(=\frac{1}{\sqrt{2}}\ln{\left|x+\frac{3}{4}+\sqrt{x^2+\frac{3}{2}x+2}\right|}+c\)

\(\int{\frac{dx}{\sqrt{3-5x-2x^2}}}\)
\(=\int{\frac{dx}{\sqrt{2\left(\frac{3}{2}-\frac{5}{2}x-x^2\right)}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\frac{3}{2}-\frac{5}{2}x-x^2}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\frac{3}{2}+\frac{25}{16}-\frac{25}{16}-\frac{5}{2}x-x^2}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\frac{24+25}{16}-\left(x^2+\frac{5}{2}x+\frac{25}{16}\right)}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{dx}{\sqrt{\frac{49}{16}-\left\{x^2+2.x.\frac{5}{2}+\left(\frac{5}{4}\right)^2\right\}}}}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(\frac{7}{4}\right)^2-\left(x+\frac{5}{4}\right)^2}}dx}\)
\(=\frac{1}{\sqrt{2}}\int{\frac{1}{\sqrt{\left(\frac{7}{4}\right)^2-t^2}}dt}\)
\(=\frac{1}{\sqrt{2}}\sin^{-1}{\left(\frac{t}{\frac{7}{4}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\sin^{-1}{\left(\frac{4t}{7}\right)}+c\)
\(=\frac{1}{\sqrt{2}}\sin^{-1}{\left\{\frac{4\left(x+\frac{5}{4}\right)}{7}\right\}}+c\) ➜Note \(\because t=x+\frac{5}{4}\)
\(=\frac{1}{\sqrt{2}}\sin^{-1}{\left(\frac{4x+5}{7}\right)}+c\)

\(\int{\frac{(x+1)}{\sqrt{4+8x-5x^2}}dx}\)
\(=\int{\frac{\frac{9}{5}-\frac{1}{10}(8-10x)}{\sqrt{4+8x-5x^2}}dx}\)
\(=\int{\frac{\frac{9}{5}-\frac{1}{10}(8-10x)}{\sqrt{4+8x-5x^2}}dx}\)
\(=\int{\frac{\frac{9}{5}}{\sqrt{4+8x-5x^2}}dx}-\int{\frac{\frac{1}{10}(8-10x)}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5}\int{\frac{1}{\sqrt{5\left(\frac{4}{5}+\frac{8}{5}x-x^2\right)}}dx}-\frac{1}{10}\int{\frac{8-10x}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\frac{4}{5}+\frac{8}{5}x-x^2}}dx}-\frac{1}{10}\int{\frac{8-10x}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\frac{4}{5}+\frac{16}{25}-\left(\frac{4}{5}\right)^2+\frac{8}{5}x-x^2}}dx}-\frac{1}{10}\int{\frac{8-10x}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\frac{20+16}{25}-\left\{x^2-\frac{8}{5}x+\left(\frac{4}{5}\right)^2\right\}}}dx}-\frac{1}{10}\int{\frac{8-10x}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\frac{36}{25}-\left\{x^2-2.x\frac{4}{5}+\left(\frac{4}{5}\right)^2\right\}}}dx}-\frac{1}{10}\int{\frac{8-10x}{\sqrt{4+8x-5x^2}}dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\left(\frac{6}{5}\right)^2-\left(x-\frac{4}{5}\right)^2}}dx}-\frac{1}{10}\int{\frac{1}{\sqrt{4+8x-5x^2}}(8-10x)dx}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\left(\frac{6}{5}\right)^2-t^2_{1}}}dt_{1}}-\frac{1}{10}\int{\frac{1}{\sqrt{t_{2}}}dt_{2}}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\left(\frac{6}{5}\right)^2-t^2_{1}}}dt_{1}}-\frac{1}{10}\int{\frac{1}{t^{\frac{1}{2}}_{2}}dt_{2}}\)
\(=\frac{9}{5\sqrt{5}}\int{\frac{1}{\sqrt{\left(\frac{6}{5}\right)^2-t^2_{1}}}dt_{1}}-\frac{1}{10}\int{t^{-\frac{1}{2}}_{2}dt_{2}}\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{t_{1}}{\frac{6}{5}}\right)}-\frac{1}{10}\frac{t^{-\frac{1}{2}+1}_{2}}{-\frac{1}{2}+1}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{5t_{1}}{6}\right)}-\frac{1}{10}\frac{t^{\frac{-1+2}{2}}_{2}}{\frac{-1+2}{2}}+c\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{5t_{1}}{6}\right)}-\frac{1}{10}\frac{t^{\frac{1}{2}}_{2}}{\frac{1}{2}}+c\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{5t_{1}}{6}\right)}-\frac{1}{10}.2t^{\frac{1}{2}}_{2}+c\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{5t_{1}}{6}\right)}-\frac{1}{5}\sqrt{t_{2}}+c\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\frac{5\left(x-\frac{4}{5}\right)}{6}}-\frac{1}{5}\sqrt{4+8x-5x^2}+c\) ➜Note \(\because t_{1}=x-\frac{4}{5}, t_{2}=4+8x-5x^2\)
\(=\frac{9}{5\sqrt{5}}\sin^{-1}{\left(\frac{5x-4}{6}\right)}-\frac{1}{5}\sqrt{4+8x-5x^2}+c\)

\(\int{\frac{dx}{(2x-3)\sqrt{3x+2}}}\)
\(=\int{\frac{1}{(2x-3)\sqrt{3x+2}}dx}\)
\(=\int{\frac{1}{\frac{2t^2-13}{3}t}\frac{2}{3}tdt}\)
\(=\frac{2}{3}\int{\frac{3}{2t^2-13}dt}\)
\(=2\int{\frac{1}{2t^2-13}dt}\)
\(=2\int{\frac{1}{2\left(t^2-\frac{13}{2}\right)}dt}\)
\(=\int{\frac{1}{t^2-\frac{13}{2}}dt}\)
\(=\int{\frac{1}{t^2-\left(\sqrt{\frac{13}{2}}\right)^2}dt}\)
\(=\frac{1}{2.\sqrt{\frac{13}{2}}}\ln{\frac{t-\sqrt{\frac{13}{2}}}{t+\sqrt{\frac{13}{2}}}}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}.\sqrt{2}.\frac{\sqrt{13}}{\sqrt{2}}}\ln{\left|\frac{t-\frac{\sqrt{13}}{\sqrt{2}}}{t+\frac{\sqrt{13}}{\sqrt{2}}}\right|}+c\) \(=\frac{1}{\sqrt{2}.\sqrt{13}}\ln{\left|\frac{\sqrt{2}t-\sqrt{13}}{\sqrt{2}t+\sqrt{13}}\right|}+c\) ➜Note লব ও হরের সহিত \(\because \sqrt{2}\) গুণ করে।
\(=\frac{1}{\sqrt{26}}\ln{\left|\frac{\sqrt{2}\sqrt{3x+2}-\sqrt{13}}{\sqrt{2}\sqrt{3x+2}+\sqrt{13}}\right|}+c\) ➜Note \(\because t=\sqrt{3x+2}\)
\(=\frac{1}{\sqrt{26}}\ln{\left|\frac{\sqrt{6x+4}-\sqrt{13}}{\sqrt{6x+4}+\sqrt{13}}\right|}+c\)

\(\int{\frac{dx}{(x-3)\sqrt{2x^2-12x+17}}}\)
\(=\int{\frac{1}{(x-3)\sqrt{2x^2-12x+17}}dx}\)
\(=\int{\frac{1}{\frac{1}{t}\sqrt{2\left(\frac{1+3t}{t}\right)^2-12\left(\frac{1+3t}{t}\right)+17}}\times{-\frac{1}{t^2}dt}}\)
\(=-\int{\frac{1}{\sqrt{2\frac{(1+3t)^2}{t^2}-\frac{12+36t}{t}+17}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\frac{1}{t}\sqrt{2(1+6t+9t^2)-12t-36t^2+17t^2}}\times{\frac{1}{t}dt}}\)
\(=-\int{\frac{1}{\sqrt{2+12t+18t^2-12t-19t^2}}dt}\)
\(=-\int{\frac{1}{\sqrt{2-t^2}}dt}\)
\(=-\int{\frac{1}{\sqrt{(\sqrt{2})^2-t^2}}dt}\)
\(=-\sin^{-1}{\left(\frac{t}{\sqrt{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\sin^{-1}{\left(\frac{\frac{1}{x-3}}{\sqrt{2}}\right)}+c\) ➜Note \(\because x-3=\frac{1}{t} \Rightarrow t=\frac{1}{x-3}\)
\(=-\sin^{-1}{\left(\frac{1}{\sqrt{2}(x-3)}\right)}+c\)

\(\int{\frac{dx}{(2x^2+a^2)\sqrt{x^2+a^2}}}\)
\(=\int{\frac{\frac{xt}{1-t^2}dt}{\left\{2\left(\frac{a^2}{t^2-1}\right)+a^2\right\}xt}}\)
\(=\int{\frac{xtdt}{\left(\frac{2a^2+a^2t^2-a^2}{t^2-1}\right)(1-t^2)xt}}\)
\(=\int{\frac{dt}{\left(\frac{2a^2+a^2t^2-a^2}{t^2-1}\right)(1-t^2)}}\)
\(=\int{\frac{xtdt}{\left\{\frac{2a^2+a^2t^2-a^2}{-(1-t^2)}\right\}(1-t^2)xt}}\)
\(=-\int{\frac{dt}{a^2t^2+a^2}}\)
\(=-\int{\frac{dt}{a^2(1+t^2)}}\)
\(=\frac{1}{a^2}\int{\left(-\frac{1}{1+t^2}\right)dt}\)
\(=\frac{1}{a^2}\cot^{-1}{t}+c\) ➜Note \(\because \int{\left(-\frac{1}{1+x^2}\right)dx}=\cot^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a^2}\cot^{-1}{\left(\frac{\sqrt{x^2+a^2}}{x}\right)}+c\) ➜Note \(\because xt=\sqrt{x^2+a^2} \Rightarrow t=\frac{\sqrt{x^2+a^2}}{x}\)
\(=\frac{1}{a^2}\tan^{-1}{\left(\frac{x}{\sqrt{x^2+a^2}}\right)}+c\)

\(\int{\frac{dx}{(x-b)^3(x-a)^2}}\)
\(=\int{\frac{\frac{x-a}{1-t}dt}{t^3(x-a)^3(x-a)^2}}\)
\(=\int{\frac{(x-a)dt}{(1-t)t^3(x-a)^5}}\)
\(=\int{\frac{dt}{(1-t)t^3(x-a)^4}}\)
\(=\int{\frac{dt}{(1-t)t^3\left(-\frac{a-b}{1-t}\right)^4}}\)
\(=\int{\frac{dt}{(1-t)t^3\times{\frac{(a-b)^4}{(1-t)^4}}}}\)
\(=\int{\frac{(1-t)^4dt}{(1-t)t^3(a-b)^4}}\)
\(=\frac{1}{(a-b)^4}\int{\frac{(1-t)^3dt}{t^3}}\)
\(=\frac{1}{(a-b)^4}\int{\frac{1-3t+3t^2-t^3}{t^3}dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(\frac{1}{t^3}-3\frac{t}{t^3}+3\frac{t^2}{t^3}-\frac{t^3}{t^3}\right)dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(\frac{1}{t^3}-3\frac{1}{t^2}+3\frac{1}{t}-1\right)dt}\)
\(=\frac{1}{(a-b)^4}\int{\left(t^{-3}-3t^{-2}+3\frac{1}{t}-1\right)dt}\)
\(=\frac{1}{(a-b)^4}\left[\int{t^{-3}dt}-3\int{t^{-2}dt}+3\int{\frac{1}{t}dt}-\int{dt}\right]\)
\(=\frac{1}{(a-b)^4}\left[\frac{t^{-3+1}}{-3+1}-3\frac{t^{-2+1}}{-2+1}+3\ln{|t|}-t\right]+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\frac{1}{x}dx}=\ln{|x|}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{(a-b)^4}\left[-\frac{1}{2t^2}+\frac{3}{t}+3\ln{|t|}-t\right]+c\)
\(=\frac{1}{(a-b)^4}\left[-\frac{1}{2\left(\frac{x-b}{x-a}\right)^2}+\frac{3}{\frac{x-b}{x-a}}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{x-b}{x-a}\right]+c\) ➜Note \(\because x-b=t(x-a)\Rightarrow \frac{x-b}{x-a}=t\)
\(=\frac{1}{(a-b)^4}\left[-\frac{(x-a)^2}{2(x-b)^2}+\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{x-b}{x-a}\right]+c\)
\(=\frac{1}{(a-b)^4}\left[\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{(x-a)^2}{2(x-b)^2}-\frac{x-b}{x-a}\right]+c\)

\(\int{\frac{dx}{9x^2+4}}\)
\(=\int{\frac{dx}{9\left(x^2+\frac{4}{9}\right)}}\)
\(=\frac{1}{9}\int{\frac{1}{x^2+\frac{4}{9}}dx}\)
\(=\frac{1}{9}\int{\frac{1}{\left(\frac{2}{3}\right)^2+x^2}dx}\)
\(=\frac{1}{9}\times{\frac{1}{\frac{2}{3}}}\tan^{-1}{\left(\frac{x}{\frac{2}{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{9}\times{\frac{3}{2}}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)
\(=\frac{1}{3}\times{\frac{1}{2}}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)
\(=\frac{1}{6}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)

\(\int{\frac{dx}{\sqrt{5-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{4\left(\frac{5}{4}-x^2\right)}}}\)
\(=\int{\frac{dx}{2\sqrt{\left(\frac{5}{4}-x^2\right)}}}\)
\(=\frac{1}{2}\int{\frac{dx}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2-x^2}}}\)
\(=\frac{1}{2}\sin^{-1}{\left(\frac{x}{\frac{\sqrt{5}}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sin^{-1}{\left(\frac{2x}{\sqrt{5}}\right)}+c\)

\(\int{\frac{1}{16x^2-9}dx}\)
\(=\int{\frac{1}{16\left(x^2-\frac{9}{16}\right)}dx}\)
\(=\frac{1}{16}\int{\frac{1}{x^2-\left(\frac{3}{4}\right)^2}dx}\)
\(=\frac{1}{16}\times{\frac{1}{2.\frac{3}{4}}}\ln{\left|\frac{x-\frac{3}{4}}{x+\frac{3}{4}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{16}\times{\frac{4}{6}}\ln{\left|\frac{\frac{4x-3}{4}}{\frac{4x+3}{4}}\right|}+c\)
\(=\frac{1}{4}\times{\frac{1}{6}}\ln{\left|\frac{\frac{4x-3}{4}}{\frac{4x+3}{4}}\right|}+c\)
\(=\frac{1}{24}\ln{\left|\frac{4x-3}{4x+2}\right|}+c\)

\(\int{\frac{dx}{16+x^2}}\)
\(=\int{\frac{dx}{4^2+x^2}}\)
\(=\frac{1}{4}\tan^{-1}{\left(\frac{x}{4}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।

\(\int{\frac{1}{16a^2+x^2}dx}\)
\(=\int{\frac{dx}{(4a)^2+x^2}}\)
\(=\frac{1}{4a}\tan^{-1}{\left(\frac{x}{4a}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।

\(\int{\frac{1}{9+x^2}dx}\)
\(=\int{\frac{dx}{3^2+x^2}}\)
\(=\frac{1}{3}\tan^{-1}{\left(\frac{x}{3}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।

\(\int{\frac{1}{9x^2+4}dx}\)
\(=\int{\frac{1}{4+9x^2}dx}\)
\(=\int{\frac{1}{9\left(\frac{4}{9}+x^2\right)}dx}\)
\(=\frac{1}{9}\int{\frac{1}{\left(\frac{2}{3}\right)^2+x^2}dx}\)
\(=\frac{1}{9}\times{\frac{1}{\frac{2}{3}}}\tan^{-1}{\left(\frac{x}{\frac{2}{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{9}\times{\frac{3}{2}}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)
\(=\frac{1}{3}\times{\frac{1}{2}}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)
\(=\frac{1}{6}\tan^{-1}{\left(\frac{3x}{2}\right)}+c\)

\(\int{\frac{dx}{\sqrt{25-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{(5)^2-x^2}}}\)
\(=\sin^{-1}{\left(\frac{x}{5}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।

\(\int{\frac{dx}{\sqrt{25-16x^2}}}\)
\(=\int{\frac{dx}{\sqrt{16\left(\frac{25}{16}-x^2\right)}}}\)
\(=\int{\frac{dx}{4\sqrt{\left(\frac{5}{4}\right)^2-x^2}}}\)
\(=\frac{1}{4}\int{\frac{dx}{\sqrt{\left(\frac{5}{4}\right)^2-x^2}}}\)
\(=\frac{1}{4}\sin^{-1}{\left(\frac{x}{\frac{5}{4}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin^{-1}{\left(\frac{4x}{5}\right)}+c\)

\(\int{\frac{dx}{\sqrt{5-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{5-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{4\left(\frac{5}{4}-x^2\right)}}}\)
\(=\int{\frac{dx}{2\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2-x^2}}}\)
\(=\frac{1}{2}\int{\frac{dx}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2-x^2}}}\)
\(=\frac{1}{2}\sin^{-1}{\left(\frac{x}{\frac{\sqrt{5}}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sin^{-1}{\left(\frac{2x}{\sqrt{5}}\right)}+c\)

\(\int{\frac{dx}{\sqrt{2-3x^2}}}\)
\(=\int{\frac{dx}{\sqrt{3\left(\frac{2}{3}-x^2\right)}}}\)
\(=\int{\frac{dx}{\sqrt{3}\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\int{\frac{dx}{\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\sin^{-1}{\left(\frac{x}{\sqrt{\frac{3}{2}}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{3}}\sin^{-1}{\left(\sqrt{\frac{3}{2}}x\right)}+c\)

\(\int{\frac{dx}{\sqrt{9-16x^2}}}\)
\(=\int{\frac{dx}{\sqrt{16\left(\frac{9}{16}-x^2\right)}}}\)
\(=\int{\frac{dx}{4\sqrt{\left(\frac{3}{4}\right)^2-x^2}}}\)
\(=\frac{1}{4}\int{\frac{dx}{\sqrt{\left(\frac{3}{4}\right)^2-x^2}}}\)
\(=\frac{1}{4}\sin^{-1}{\left(\frac{x}{\frac{3}{4}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin^{-1}{\left(\frac{4x}{3}\right)}+c\)

\(\int{\frac{dx}{\sqrt{12-16x^2}}}\)
\(=\int{\frac{dx}{\sqrt{16\left(\frac{12}{16}-x^2\right)}}}\)
\(=\int{\frac{dx}{\sqrt{16\left(\frac{3}{4}-x^2\right)}}}\)
\(=\int{\frac{dx}{4\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2-x^2}}}\)
\(=\frac{1}{4}\int{\frac{dx}{\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2-x^2}}}\)
\(=\frac{1}{4}\sin^{-1}{\left(\frac{x}{\frac{\sqrt{3}}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin^{-1}{\left(\frac{2x}{\sqrt{3}}\right)}+c\)

\(\int{\frac{dx}{16-4x^2}}\)
\(=\int{\frac{dx}{4\left(4-x^2\right)}}\)
\(=\frac{1}{4}\int{\frac{dx}{(2)^2-x^2}}\)
\(=\frac{1}{4}\times{\frac{1}{2.2}}\ln{\left|\frac{2+x}{2-x}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\times{\frac{1}{4}}\ln{\left|\frac{2+x}{2-x}\right|}+c\)
\(=\frac{1}{16}\ln{\left|\frac{2+x}{2-x}\right|}+c\)

\(\int{\frac{dx}{9-4x^2}}\)
\(=\int{\frac{dx}{4\left(\frac{9}{4}-x^2\right)}}\)
\(=\frac{1}{4}\int{\frac{dx}{\left(\frac{3}{2}\right)^2-x^2}}\)
\(=\frac{1}{4}\times{\frac{1}{2.\frac{3}{2}}}\ln{\left|\frac{\frac{3}{2}+x}{\frac{3}{2}-x}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\times{\frac{1}{3}}\ln{\left|\frac{\frac{3+2x}{2}}{\frac{3-2x}{2}}\right|}+c\)
\(=\frac{1}{12}\ln{\left|\frac{3+2x}{3-2x}\right|}+c\)

\(\int{\sqrt{1-9x^2}dx}\)
\(=\int{\sqrt{9\left(\frac{1}{9}-x^2\right)}dx}\)
\(=\int{3\sqrt{\left(\frac{1}{3}\right)^2-x^2}dx}\)
\(=3\int{\sqrt{\left(\frac{1}{3}\right)^2-x^2}dx}\)
\(=3\frac{x\sqrt{\left(\frac{1}{3}\right)^2-x^2}}{2}+3\frac{\left(\frac{1}{3}\right)^2}{2}\sin^{-1}\left(\frac{x}{\frac{1}{3}}\right)+c\) ➜Note \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=3\frac{x\sqrt{\frac{1}{9}-x^2}}{2}+3\frac{\frac{1}{9}}{2}\sin^{-1}\left(\frac{x}{\frac{1}{3}}\right)+c\)
\(=3\frac{x\sqrt{\frac{1-9x^2}{9}}}{2}+3\frac{1}{6}\sin^{-1}(3x)+c\)
\(=\frac{3x\frac{\sqrt{1-9x^2}}{3}}{2}+\frac{1}{6}\sin^{-1}(3x)+c\)
\(=\frac{x\sqrt{1-9x^2}}{2}+\frac{1}{6}\sin^{-1}(3x)+c\)

\(\int{\frac{1}{2x^2+x+1}dx}\)
\(=\int{\frac{1}{2\left(x^2+\frac{x}{2}+\frac{1}{2}\right)}dx}\)
\(=\frac{1}{2}\int{\frac{1}{x^2+\frac{x}{2}+\frac{1}{2}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2-\frac{1}{16}+\frac{1}{2}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2+\frac{1}{2}-\frac{1}{16}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2+\frac{8-1}{16}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\left(x+\frac{1}{4}\right)^2+\frac{7}{16}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{7}{16}+\left(x+\frac{1}{4}\right)^2}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\left(\frac{\sqrt{7}}{4}\right)^2+t^2}dt}\)
\(=\frac{1}{2}.\frac{1}{\frac{\sqrt{7}}{4}}\tan^{-1}{\frac{t}{\frac{\sqrt{7}}{4}}}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}.\frac{4}{\sqrt{7}}\tan^{-1}{\left(\frac{4t}{\sqrt{7}}\right)}+c\)
\(=\frac{2}{\sqrt{7}}\tan^{-1}{\left(\frac{4\left(x+\frac{1}{4}\right)}{\sqrt{7}}\right)}+c\) ➜Note \(\because t=x+\frac{1}{4}\)
\(=\frac{2}{\sqrt{7}}\tan^{-1}{\left(\frac{4x+1}{\sqrt{7}}\right)}+c\)

\(\int{\frac{1}{x^2-x+1}dx}\)
\(=\int{\frac{1}{x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}+1}dx}\)
\(=\int{\frac{1}{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+1}dx}\)
\(=\int{\frac{1}{\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}}dx}\)
\(=\int{\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{4-1}{4}}dx}\)
\(=\int{\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx}\)
\(=\int{\frac{1}{\frac{3}{4}+\left(x-\frac{1}{2}\right)^2}dx}\)
\(=\int{\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2+t^2}dt}\)
\(=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}{\frac{t}{\frac{\sqrt{3}}{2}}}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\frac{2t}{\sqrt{3}}}+c\)
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\frac{2\left(x-\frac{1}{2}\right)}{\sqrt{3}}}+c\) ➜Note \(\because t=x-\frac{1}{2}\)
\(=\frac{2}{\sqrt{3}}\tan^{-1}{\left(\frac{2x-1}{\sqrt{3}}\right)}+c\)

\(\int{\frac{dx}{\sqrt{x(4-x)}}}\)
\(=\int{\frac{dx}{\sqrt{4x-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{4-4+4x-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{4-(x^2-4x+4)}}}\)
\(=\int{\frac{dx}{\sqrt{4-(x^2-2.x.2+2^2)}}}\)
\(=\int{\frac{1}{\sqrt{2^2-(x-2)^2}}dx}\)
\(=\int{\frac{1}{\sqrt{2^2-t^2}}dt}\)
\(=\sin^{-1}{\left(\frac{t}{2}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{\left(\frac{x-2}{2}\right)}+c\) ➜Note \(\because t=x-2\)

\(\int{\frac{dx}{\sqrt{2ax-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{2ax-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{a^2-a^2+2ax-x^2}}}\)
\(=\int{\frac{dx}{\sqrt{a^2-(x^2-2ax+a^2)}}}\)
\(=\int{\frac{dx}{\sqrt{a^2-(x^2-2.x.a+a^2)}}}\)
\(=\int{\frac{1}{\sqrt{a^2-(x-a)^2}}dx}\)
\(=\int{\frac{1}{\sqrt{a^2-t^2}}dt}\)
\(=\sin^{-1}{\left(\frac{t}{a}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{\left(\frac{x-a}{a}\right)}+c\) ➜Note \(\because t=x-2\)

\(\int{\frac{dx}{\sqrt{15-4x-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{16-1-4x-4x^2}}}\)
\(=\int{\frac{dx}{\sqrt{16-(4x^2+4x+1)}}}\)
\(=\int{\frac{dx}{\sqrt{16-\{(2x)^2+2.2x.1+1^2\}}}}\)
\(=\int{\frac{1}{\sqrt{16-(2x+1)^2}}dx}\)
\(=\int{\frac{1}{\sqrt{16-t^2}}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\sqrt{4^2-t^2}}dt}\)
\(=\frac{1}{2}\sin^{-1}{\left(\frac{t}{4}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sin^{-1}{\left(\frac{2x+1}{4}\right)}+c\) ➜Note \(\because t=2x+1\)

\(\int{\frac{1}{5+4x-x^2}dx}\)
\(=\int{\frac{1}{9-4+4x-x^2}dx}\)
\(=\int{\frac{dx}{9-(x^2-4x+4)}}\)
\(=\int{\frac{dx}{9-(x^2-2.x.2+2^2)}}\)
\(=\int{\frac{1}{9-(x-2)^2}dx}\)
\(=\int{\frac{1}{9-t^2}dt}\)
\(=\int{\frac{1}{3^2-t^2}dt}\)
\(=\frac{1}{2.3}\ln{\left|\frac{3+t}{3-t}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{6}\ln{\left|\frac{3+x-2}{3-x+2}\right|}+c\) ➜Note \(\because t=x-2\)
\(=\frac{1}{6}\ln{\left|\frac{x+1}{5-x}\right|}+c\)

\(\int{\frac{1}{\sqrt{x^2+4x+13}}dx}\)
\(=\int{\frac{1}{\sqrt{x^2+4x+4+9}}dx}\)
\(=\int{\frac{1}{\sqrt{x^2+2.x.2+2^2+9}}dx}\)
\(=\int{\frac{1}{\sqrt{(x+2)^2+9}}dx}\)
\(=\int{\frac{1}{\sqrt{t^2+3^2}}dt}\)
\(=\int{\frac{1}{\sqrt{3^2+t^2}}dt}\)
\(=\ln{|\sqrt{3^2+t^2}+t|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2+x^2}}dx}=\ln{|\sqrt{a^2+x^2}+x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sqrt{9+(x+2)^2}+(x+2)|}+c\) ➜Note \(\because t=x+2\)
\(=\ln{|\sqrt{9+x^2+4x+4}+(x+2)|}+c\)
\(=\ln{|\sqrt{x^2+4x+13}+(x+2)|}+c\)
\(=\ln{\left|(x+2)+\sqrt{x^2+4x+13}\right|}+c\)

\(\int{\frac{dx}{\sqrt{2x+x^2}}}\)
\(=\int{\frac{dx}{\sqrt{x^2+2x+1-1}}}\)
\(=\int{\frac{1}{\sqrt{x^2+2.x.1+1^2-1}}dx}\)
\(=\int{\frac{1}{\sqrt{(x+1)^2-1}}dx}\)
\(=\int{\frac{1}{\sqrt{t^2-1^2}}dt}\)
\(=\ln{|\sqrt{t^2-1^2}+t|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|\sqrt{x^2-a^2}+x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sqrt{(x+1)^2-1}+(x+1)|}+c\) ➜Note \(\because t=x+1\)
\(=\ln{|\sqrt{x^2+2x+1-1}+(x+1)|}+c\)
\(=\ln{|\sqrt{x^2+2x}+(x+1)|}+c\)
\(=\ln{|(x+1)+\sqrt{2x+x^2}|}+c\)

\(\int{\sqrt{2ax-x^2}dx}\)
\(=\int{\sqrt{a^2-a^2+2ax-x^2}dx}\)
\(=\int{\sqrt{a^2-(x^2-2ax+a^2)}dx}\)
\(=\int{\sqrt{a^2-(x-a)^2}dx}\)
\(=\int{\sqrt{a^2-(x-a)^2}dx}\)
\(=\int{\sqrt{a^2-t^2}dt}\)
\(=\frac{t\sqrt{a^2-t^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{t}{a}\right)}+c\) ➜Note \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x-a)\sqrt{a^2-(x-a)^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x-a}{a}\right)}+c\) ➜Note \(\because t=x-a\)
\(=\frac{(x-a)\sqrt{a^2-x^2+2ax-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x-a}{a}\right)}+c\)
\(=\frac{(x-a)\sqrt{2ax-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x-a}{a}\right)}+c\)

\(\int{\frac{1}{4x^2+9}dx}\)
\(=\int{\frac{1}{9+4x^2}dx}\)
\(=\int{\frac{1}{4\left(\frac{9}{4}+x^2\right)}dx}\)
\(=\frac{1}{4}\int{\frac{1}{\left(\frac{3}{2}\right)^2+x^2}dx}\)
\(=\frac{1}{4}\times{\frac{1}{\frac{3}{2}}}\tan^{-1}{\left(\frac{x}{\frac{3}{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\times{\frac{2}{3}}\tan^{-1}{\left(\frac{2x}{3}\right)}+c\)
\(=\frac{1}{2}\times{\frac{1}{3}}\tan^{-1}{\left(\frac{2x}{3}\right)}+c\)
\(=\frac{1}{6}\tan^{-1}{\left(\frac{2x}{3}\right)}+c\)

\(\int{\frac{1}{x^2+4x+13}dx}\)
\(=\int{\frac{1}{x^2+4x+4+9}dx}\)
\(=\int{\frac{1}{x^2+2.x.2+2^2+9}dx}\)
\(=\int{\frac{1}{(x+2)^2+9}dx}\)
\(=\int{\frac{1}{t^2+3^2}dt}\)
\(=\int{\frac{1}{3^2+t^2}dt}\)
\(=\frac{1}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan^{-1}{\left(\frac{x+2}{3}\right)}+c\) ➜Note \(\because t=x+2\)

\(\int{\frac{dx}{\sqrt{12x-9x^2}}}\)
\(=\int{\frac{dx}{\sqrt{9\left(\frac{4}{3}x-x^2\right)}}}\)
\(=\int{\frac{dx}{3\sqrt{\frac{4}{9}-\left(\frac{2}{3}\right)^2+\frac{4}{3}x-x^2}}}\)
\(=\frac{1}{3}\int{\frac{dx}{\sqrt{\frac{4}{9}-\left\{x^2-\frac{4}{3}x+\left(\frac{2}{3}\right)^2\right\}}}}\)
\(=\frac{1}{3}\int{\frac{dx}{\sqrt{\frac{4}{9}-\left\{x^2-2.x.\frac{2}{3}x+\left(\frac{2}{3}\right)^2\right\}}}}\)
\(=\frac{1}{3}\int{\frac{dx}{\sqrt{\left(\frac{2}{3}\right)^2-\left(x-\frac{2}{3}\right)^2}}}\)
\(=\frac{1}{3}\int{\frac{dx}{\sqrt{\left(\frac{2}{3}\right)^2-t^2}}}\)
\(=\frac{1}{3}\sin^{-1}{\left(\frac{t}{\frac{2}{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\sin^{-1}{\left(\frac{3t}{2}\right)}+c\)
\(=\frac{1}{3}\sin^{-1}{\left\{\frac{3\left(x-\frac{2}{3}\right)}{2}\right\}}+c\) ➜Note \(\because t=x-\frac{2}{3}\)
\(=\frac{1}{3}\sin^{-1}{\left(\frac{3x-2}{2}\right)}+c\)

\(\int{\frac{dx}{x^2+x}}\)
\(=\int{\frac{dx}{x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}}}\)
\(=\int{\frac{1}{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}dx}\)
\(=\int{\frac{1}{t^2-\left(\frac{1}{2}\right)^2}dt}\)
\(=\frac{1}{2.\frac{1}{2}}\ln{\left|\frac{t-\frac{1}{2}}{t+\frac{1}{2}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{1}\ln{\left|\frac{2t-1}{2t+1}\right|}+c\) ➜Note লব ও হরের সহিত \(2\) গুণ করে।
\(=\ln{\left|\frac{2t-1}{2t+1}\right|}+c\)
\(=\ln{\left|\frac{2\left(x+\frac{1}{2}\right)-1}{2\left(x+\frac{1}{2}\right)+1}\right|}+c\) ➜Note \(\because t=x+\frac{1}{2}\)
\(=\ln{\left|\frac{2x+1-1}{2x+1+1}\right|}+c\)
\(=\ln{\left|\frac{2x}{2x+2}\right|}+c\)
\(=\ln{\left|\frac{2x}{2(x+1)}\right|}+c\)
\(=\ln{\left|\frac{x}{x+1}\right|}+c\)

\(\int{\sqrt{16-9x^2}dx}\)
\(=\int{\sqrt{9\left(\frac{16}{9}-x^2\right)}dx}\)
\(=\int{3\sqrt{\left(\frac{4}{3}\right)^2-x^2}dx}\)
\(=3\int{\sqrt{\left(\frac{4}{3}\right)^2-x^2}dx}\)
\(=3\frac{x\sqrt{\left(\frac{4}{3}\right)^2-x^2}}{2}+3\frac{\left(\frac{4}{3}\right)^2}{2}\sin^{-1}\left(\frac{x}{\frac{4}{3}}\right)+c\) ➜Note \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=3\frac{x\sqrt{\frac{16}{9}-x^2}}{2}+3\frac{\frac{16}{9}}{2}\sin^{-1}\left(\frac{x}{\frac{4}{3}}\right)+c\)
\(=3\frac{x\sqrt{\frac{16-9x^2}{9}}}{2}+3\frac{16}{18}\sin^{-1}{\left(\frac{3x}{4}\right)}+c\)
\(=\frac{3x\frac{\sqrt{16-9x^2}}{3}}{2}+\frac{16}{6}\sin^{-1}{\left(\frac{3x}{4}\right)}+c\)
\(=\frac{x\sqrt{16-9x^2}}{2}+\frac{8}{3}\sin^{-1}{\left(\frac{3x}{4}\right)}+c\)

\(L.S=\int{\sqrt{a^2-x^2}dx}\)
\(=\int{\sqrt{a^2-a^2\sin^2{\theta}}.a\cos{\theta}d\theta}\)
\(=\int{\sqrt{a^2(1-\sin^2{\theta})}.a\cos{\theta}d\theta}\)
\(=\int{\sqrt{a^2\cos^2{\theta}}.a\cos{\theta}d\theta}\) ➜Note \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{a\cos{\theta}.a\cos{\theta}d\theta}\)
\(=a^2\int{\cos^2{\theta}d\theta}\)
\(=\frac{a^2}{2}\int{2\cos^2{\theta}d\theta}\)
\(=\frac{a^2}{2}\int{(1+\cos{2\theta})d\theta}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{a^2}{2}\left\{\theta+\frac{\sin{2\theta}}{2}\right\}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{\sin{ax}}{a} \) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a^2}{2}\left\{\theta+\frac{2\sin{\theta}\cos{\theta}}{2}\right\}+c\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=\frac{a^2}{2}\left\{\theta+\sin{\theta}\cos{\theta}\right\}+c\)
\(=\frac{a^2}{2}\left\{\theta+\sin{\theta}\sqrt{1-\sin^2{\theta}}\right\}+c\) ➜Note \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}\sqrt{1-\left(\frac{x}{a}\right)^2}\right\}+c\) ➜Note \(\because \theta=\sin^{-1}\left(\frac{x}{a}\right), \sin{\theta}=\frac{x}{a}\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}\sqrt{\frac{a^2-x^2}{a^2}}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x}{a}.\frac{\sqrt{a^2-x^2}}{a}\right\}+c\)
\(=\frac{a^2}{2}\left\{\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x\sqrt{a^2-x^2}}{a^2}\right\}+c\)
\(=\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x\sqrt{a^2-x^2}}{2}+c\)
\(=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}+c\)
\(=\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}+\frac{x\sqrt{a^2-x^2}}{2}+c\)

\(\int{\frac{xdx}{x^4+1}}\)
\(=\int{\frac{1}{1+(x^2)^2}xdx}\)
\(=\int{\frac{1}{1+t^2}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{1+t^2}dt}\)
\(=\frac{1}{2}\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\tan^{-1}{(x^2)}+c\) ➜Note\(\because t=x^2\)

\(\int{\frac{xdx}{x^4-4}}\)
\(=\int{\frac{1}{(x^2)^2-4}xdx}\)
\(=\int{\frac{1}{t^2-4}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t^2-2^2}dt}\)
\(=\frac{1}{2}.\frac{1}{2.2}\ln{\left|\frac{t-2}{t+2}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}.\frac{1}{4}\ln{\left|\frac{x^2-2}{x^2+2}\right|}+c\) ➜Note\(\because t=x^2\)
\(=\frac{1}{8}\ln{\left|\frac{x^2-2}{x^2+2}\right|}+c\)

\(\int{\frac{xdx}{x^4+3}}\)
\(=\int{\frac{1}{(x^2)^2+3}xdx}\)
\(=\int{\frac{1}{t^2+3}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t^2+(\sqrt{3})^2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{(\sqrt{3})^2+t^2}dt}\)
\(=\frac{1}{2}.\frac{1}{\sqrt{3}}\tan^{-1}{\left(\frac{t}{\sqrt{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2\sqrt{3}}\tan^{-1}{\left(\frac{x^2}{\sqrt{3}}\right)}+c\) ➜Note\(\because t=x^2\)

\(\int{\frac{3x^2}{1+x^6}dx}\)
\(=\int{\frac{1}{1+(x^3)^2}3x^2dx}\)
\(=\int{\frac{1}{1+t^2}dt}\)
\(=\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan^{-1}{(x^3)}+c\) ➜Note\(\because t=x^3\)

\(\int{\frac{5e^{2x}}{1+e^{4x}}dx}\)
\(=5\int{\frac{1}{1+\left(e^{2x}\right)^2}e^{2x}dx}\)
\(=5\int{\frac{1}{1+t^2}\frac{1}{2}dt}\)
\(=\frac{5}{2}\int{\frac{1}{1+t^2}dt}\)
\(=\frac{5}{2}\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{5}{2}\tan^{-1}{\left(e^{2x}\right)}+c\) ➜Note\(\because t=e^{2x}\)

\(\int{\frac{1}{e^x+e^{-x}}dx}\)
\(=\int{\frac{e^x}{e^x(e^x+e^{-x})}dx}\) ➜Note লব ও হরের সহিত \(e^{x}\) গুণ করে।
\(=\int{\frac{e^x}{(e^x)^2+e^{-x+x}}dx}\)
\(=\int{\frac{e^x}{(e^x)^2+e^{0}}dx}\)
\(=\int{\frac{1}{(e^x)^2+1}e^xdx}\)
\(=\int{\frac{1}{t^2+1}dt}\)
\(=\int{\frac{1}{1+t^2}dt}\)
\(=\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan^{-1}{\left(e^{x}\right)}+c\) ➜Note\(\because t=e^{x}\)

\(\int{\frac{x^2}{e^{x^3}+e^{-x^3}}dx}\)
\(=\int{\frac{x^2e^{x^3}}{e^{x^3}\left(e^{x^3}+e^{-x^3}\right)}dx}\) ➜Note লব ও হরের সহিত \(e^{x^3}\) গুণ করে।
\(=\int{\frac{x^2e^{x^3}}{\left(e^{x^3}\right)^2+e^{-x^3+x^3}}dx}\)
\(=\int{\frac{x^2e^{x^3}}{\left(e^{x^3}\right)^2+e^{0}}dx}\)
\(=\int{\frac{1}{\left(e^{x^3}\right)^2+1}x^2e^{x^3}dx}\)
\(=\int{\frac{1}{t^2+1}\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\frac{1}{1+t^2}dt}\)
\(=\frac{1}{3}\tan^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan^{-1}{\left(e^{x^3}\right)}+c\) ➜Note\(\because t=e^{x^3}\)

\(\int{\frac{\cos{2x}dx}{\sin^2{2x}+8}}\)
\(=\int{\frac{1}{\sin^2{2x}+8}\cos{2x}dx}\)
\(=\int{\frac{1}{t^2+8}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{8+t^2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{(2\sqrt{2})^2+t^2}dt}\)
\(=\frac{1}{2}.\frac{1}{2\sqrt{2}}\tan^{-1}{\left(\frac{t}{2\sqrt{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4\sqrt{2}}\tan^{-1}{\left(\frac{\sin{2x}}{2\sqrt{2}}\right)}+c\) ➜Note\(\because t=\sin{2x}\)
\(=\frac{\sqrt{2}}{8}\tan^{-1}{\left(\frac{\sin{2x}}{2\sqrt{2}}\right)}+c\)

\(\int{\frac{d\theta}{1+3\cos^2{\theta}}}\)
\(=\int{\frac{d\theta}{\cos^2{\theta}\left(\frac{1}{\cos^2{\theta}}+3\right)}}\)
\(=\int{\frac{\frac{1}{\cos^2{\theta}}d\theta}{\frac{1}{\cos^2{\theta}}+3}}\)
\(=\int{\frac{\sec^2{\theta}d\theta}{\sec^2{\theta}+3}}\) ➜Note\(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int{\frac{\sec^2{\theta}d\theta}{1+\tan^2{\theta}+3}}\) ➜Note\(\because \sec^2{A}=1+\tan^2{A}\)
\(=\int{\frac{1}{4+\tan^2{\theta}}\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{4+t^2}dt}\)
\(=\int{\frac{1}{2^2+t^2}dt}\)
\(=\frac{1}{2}\tan^{-1}{\left(\frac{t}{2}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\tan^{-1}{\left(\frac{\tan{\theta}}{2}\right)}+c\) ➜Note\(\because t=\tan{\theta}\)
\(=\frac{1}{2}\tan^{-1}{\left(\frac{1}{2}\tan{\theta}\right)}+c\)

\(\int{\frac{dx}{1+\cos^2{x}}}\)
\(=\int{\frac{dx}{\cos^2{x}\left(\frac{1}{\cos^2{x}}+1\right)}}\)
\(=\int{\frac{\frac{1}{\cos^2{x}}dx}{\frac{1}{\cos^2{x}}+1}}\)
\(=\int{\frac{\sec^2{x}dx}{\sec^2{x}+1}}\) ➜Note\(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int{\frac{\sec^2{x}dx}{1+\tan^2{x}+1}}\) ➜Note\(\because \sec^2{A}=1+\tan^2{A}\)
\(=\int{\frac{1}{2+\tan^2{x}}\sec^2{x}dx}\)
\(=\int{\frac{1}{2+t^2}dt}\)
\(=\int{\frac{1}{(\sqrt{2})^2+t^2}dt}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{t}{\sqrt{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{\tan{x}}{\sqrt{2}}\right)}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{x^2+1}{x^4+1}dx}\)
\(=\int{\frac{x^2\left(1+\frac{1}{x^2}\right)}{x^2\left(x^2+\frac{1}{x^2}\right)}dx}\)
\(=\int{\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx}\)
\(=\int{\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2.x.\frac{1}{x}}dx}\)
\(=\int{\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}dx}\)
\(=\int{\frac{1}{\left(x-\frac{1}{x}\right)^2+2}\left(1+\frac{1}{x^2}\right)dx}\)
\(=\int{\frac{1}{t^2+2}dt}\)
\(=\int{\frac{1}{2+t^2}dt}\)
\(=\int{\frac{1}{(\sqrt{2})^2+t^2}dt}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{t}{\sqrt{2}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)}+c\) ➜Note\(\because t=x-\frac{1}{x}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{\frac{x^2-1}{x}}{\sqrt{2}}\right)}+c\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{x^2-1}{\sqrt{2}x}\right)}+c\)

\(\int{\frac{\sec^2{x}dx}{\sqrt{1-\tan^2{x}}}}\)
\(=\int{\frac{1}{\sqrt{1-\tan^2{x}}}\sec^2{x}dx}\)
\(=\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{(\tan{x})}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{x^2dx}{\sqrt{1-x^6}}}\)
\(=\int{\frac{x^2dx}{\sqrt{1-(x^3)^2}}}\)
\(=\int{\frac{1}{\sqrt{1-(x^3)^2}}x^2dx}\)
\(=\int{\frac{1}{\sqrt{1-t^2}}\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=\frac{1}{3}\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\sin^{-1}{(x^3)}+c\) ➜Note\(\because t=x^3\)

\(\int{\frac{x^2-1}{x^4+1}dx}\)
\(=\int{\frac{x^2\left(1-\frac{1}{x^2}\right)}{x^2\left(x^2+\frac{1}{x^2}\right)}dx}\)
\(=\int{\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx}\)
\(=\int{\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2.x.\frac{1}{x}}dx}\)
\(=\int{\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2}dx}\)
\(=\int{\frac{1}{\left(x+\frac{1}{x}\right)^2-2}\left(1-\frac{1}{x^2}\right)dx}\)
\(=\int{\frac{1}{t^2-2}dt}\)
\(=\int{\frac{1}{t^2-(\sqrt{2})^2}dt}\)
\(=\frac{1}{2\sqrt{2}}\ln{\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2\sqrt{2}}\ln{\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|}+c\) ➜Note\(\because t=x+\frac{1}{x}\)
\(=\frac{1}{2\sqrt{2}}\ln{\left|\frac{x^2+1-\sqrt{2}x}{x^2+1+\sqrt{2}x}\right|}+c\) ➜Note লব ও হরের সহিত \(x\) গুণ করে।

\(\int{\frac{1}{e^x-e^{-x}}dx}\)
\(=\int{\frac{e^{x}}{e^{x}\left(e^x-e^{-x}\right)}dx}\) ➜Note লব ও হরের সহিত \(e^x\) গুণ করে।
\(=\int{\frac{e^{x}}{\left(e^x\right)^2-e^{-x+x}}dx}\) \(=\int{\frac{e^{x}}{\left(e^x\right)^2-e^{0}}dx}\) \(=\int{\frac{1}{t^2-1}dt}\) \(=\frac{1}{2.1}\ln{\left|\frac{t-1}{t+1}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\ln{\left|\frac{e^x-1}{e^x+1}\right|}+c\) ➜Note\(\because t=e^x\)

\(\int{\frac{x^2}{e^{x^3}-e^{-x^3}}dx}\)
\(=\int{\frac{x^2e^{x^3}}{e^{x^3}\left(e^{x^3}-e^{-x^3}\right)}dx}\) ➜Note লব ও হরের সহিত \(e^{x^3}\) গুণ করে।
\(=\int{\frac{x^2e^{x^3}}{\left(e^{x^3}\right)^2-e^{-x^3+x^3}}dx}\)
\(=\int{\frac{x^2e^{x^3}}{\left(e^{x^3}\right)^2-e^{0}}dx}\)
\(=\int{\frac{1}{\left(e^{x^3}\right)^2-1}x^2e^{x^3}dx}\)
\(=\int{\frac{1}{t^2-1}\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\frac{1}{t^2-1}dt}\)
\(=\frac{1}{3}.\frac{1}{2.1}\ln{\left|\frac{t-1}{t+1}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}.\frac{1}{2}\ln{\left|\frac{e^{x^3}-1}{e^{x^3}+1}\right|}+c\) ➜Note\(\because t=e^{x^3}\)
\(=\frac{1}{6}\ln{\left|\frac{e^{x^3}-1}{e^{x^3}+1}\right|}+c\)

\(\int{\frac{\cos{x}dx}{3+\cos^2{x}}}\)
\(=\int{\frac{\cos{x}dx}{3+1-\sin^2{x}}}\) ➜Note \(\because \cos^2{A}=1-\sin^2{A}\)
\(=\int{\frac{\cos{x}dx}{4-\sin^2{x}}}\)
\(=\int{\frac{1}{4-\sin^2{x}}\cos{x}dx}\)
\(=\int{\frac{1}{4-t^2}dt}\)
\(=\int{\frac{1}{2^2-t^2}dt}\)
\(=\frac{1}{2.2}\ln{\left|\frac{2+t}{2-t}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\ln{\left|\frac{2+\sin{x}}{2-\sin{x}}\right|}+c\) ➜Note\(\because t=\sin{x}\)

\(\int{\frac{e^x}{\sqrt{e^{2x}+1}}dx}\)
\(=\int{\frac{e^x}{\sqrt{\left(e^{x}\right)^2+1}}dx}\)
\(=\int{\frac{1}{\sqrt{\left(e^{x}\right)^2+1}}e^xdx}\)
\(=\int{\frac{1}{\sqrt{t^2+1}}dt}\)
\(=\int{\frac{1}{\sqrt{1^2+t^2}}dt}\)
\(=\ln{|\sqrt{1^2+t^2}+t|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2+x^2}}dx}=\ln{|\sqrt{a^2+x^2}+x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sqrt{1+(e^x)^2}+e^x|}+c\) ➜Note\(\because t=e^{x}\)
\(=\ln{|\sqrt{1+e^{2x}}+e^x|}+c\)
\(=\ln{|e^x+\sqrt{e^{2x}+1}|}+c\)

\(\int{\frac{dx}{(a^2+x^2)^{\frac{3}{2}}}}\)
\(=\int{\frac{1}{(a^2+x^2)^{\frac{3}{2}}}dx}\)
\(=\int{\frac{1}{(a^2+a^2\tan^2{\theta})^{\frac{3}{2}}}a\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\{a^2(1+\tan^2{\theta})\}^{\frac{3}{2}}}a\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\{a^2\sec^2{\theta}\}^{\frac{3}{2}}}a\sec^2{\theta}d\theta}\) ➜Note\(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{a^3\sec^3{\theta}}a\sec^2{\theta}d\theta}\)
\(=\frac{1}{a^2}\int{\frac{1}{\sec{\theta}}d\theta}\)
\(=\frac{1}{a^2}\int{\cos{\theta}d\theta}\)
\(=\frac{1}{a^2}\sin{\theta}+c\) ➜Note \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a^2}\sin{\tan^{-1}{\left(\frac{x}{a}\right)}}+c\) ➜Note\(\because \theta=\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{a^2}\sin{\sin^{-1}{\left(\frac{x}{\sqrt{a^2+x^2}}\right)}}+c\) ➜Note\(\because \tan^{-1}{A}=\sin^{-1}{\frac{A}{\sqrt{1+A^2}}}\)
\(=\frac{1}{a^2}\left(\frac{x}{\sqrt{a^2+x^2}}\right)+c\)
\(=\frac{x}{a^2\sqrt{a^2+x^2}}+c\)

\(\int{\frac{dx}{(x^2+9)^{2}}}\)
\(=\int{\frac{dx}{(9+x^2)^{2}}}\)
\(=\int{\frac{1}{(9+9\tan^2{\theta})^{2}}3\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\{9(1+\tan^2{\theta})\}^{2}}3\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\{9\sec^2{\theta}\}^{2}}3\sec^2{\theta}d\theta}\) ➜Note\(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{81\sec^4{\theta}}3\sec^2{\theta}d\theta}\)
\(=\frac{1}{27}\int{\frac{1}{\sec^2{\theta}}d\theta}\)
\(=\frac{1}{27}\int{\cos^2{\theta}d\theta}\)
\(=\frac{1}{54}\int{2\cos^2{\theta}d\theta}\)
\(=\frac{1}{54}\int{(1+\cos{2\theta})d\theta}\)
\(=\frac{1}{54}\int{d\theta}+\frac{1}{54}\int{\cos{2\theta}d\theta}\)
\(=\frac{1}{54}\theta+\frac{1}{54}.\frac{1}{2}\sin{2\theta}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{54}\theta+\frac{1}{108}\sin{2\theta}+c\)
\(=\frac{1}{54}\tan^{-1}{\left(\frac{x}{3}\right)}+\frac{1}{108}\sin{\left\{2\tan^{-1}{\left(\frac{x}{3}\right)}\right\}}+c\) ➜Note\(\because \theta=\tan^{-1}{\left(\frac{x}{3}\right)}\)
\(=\frac{1}{108}\left[2\tan^{-1}{\left(\frac{x}{3}\right)}+\sin{\left\{2\tan^{-1}{\left(\frac{x}{3}\right)}\right\}}\right]+c\)

\(\int{\frac{dx}{x\sqrt{x^2+1}}}\)
\(=\int{\frac{1}{x\sqrt{1+x^2}}dx}\)
\(=\int{\frac{1}{\tan{\theta}\sqrt{1+\tan^2{\theta}}}\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\tan{\theta}\sqrt{\sec^2{\theta}}}\sec^2{\theta}d\theta}\) ➜Note\(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{\tan{\theta}\sec{\theta}}\sec^2{\theta}d\theta}\)
\(=\int{\frac{1}{\tan{\theta}}\sec{\theta}d\theta}\)
\(=\int{\frac{1}{\frac{\sin{\theta}}{\cos{\theta}}}\times{\frac{1}{\cos{\theta}}}d\theta}\)
\(=\int{\frac{\cos{\theta}}{\sin{\theta}}\times{\frac{1}{\cos{\theta}}}d\theta}\)
\(=\int{\frac{1}{\sin{\theta}}d\theta}\)
\(=\int{cosec \ {\theta}d\theta}\)
\(=\ln{|cosec \ {\theta}-\cot{\theta}|}+c\) ➜Note \(\because \int{cosec \ {x}dx}=\ln{|cosec \ {x}-\cot{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|cosec \ {(\tan^{-1}{x})}-\cot{(\tan^{-1}{x})}|}+c\) ➜Note\(\because \theta=\tan^{-1}{x}\)
\(=\ln{|cosec \ {\left(cosec^{-1}{\frac{\sqrt{1+x^2}}{x}}\right)}-\cot{\left(\cot^{-1}{\frac{}{x}}\right)}|}+c\) ➜Note\(\because \tan^{-1}{A}=\cot^{-1}{\frac{1}{A}}=cosec^{-1}{\frac{\sqrt{1+A^2}}{A}}\)
\(=\ln{\left|\frac{\sqrt{1+x^2}}{x}-\frac{1}{x}\right|}+c\)
\(=\ln{\left|\frac{\sqrt{1+x^2}-1}{x}\right|}+c\)

\(\int{\frac{dx}{a^2\sin^2{x}+b^2\cos^2{x}}}\)
\(=\int{\frac{dx}{a^2\cos^2{x}\left(\frac{\sin^2{x}}{\cos^2{x}}+\frac{b^2}{a^2}\right)}}\)
\(=\frac{1}{a^2}\int{\frac{\sec^2{x}dx}{\tan^2{x}+\frac{b^2}{a^2}}}\)
\(=\frac{1}{a^2}\int{\frac{dt}{t^2+\frac{b^2}{a^2}}}\)
\(=\frac{1}{a^2}\int{\frac{dt}{\frac{b^2}{a^2}+t^2}}\)
\(=\frac{1}{a^2}\int{\frac{dt}{\left(\frac{b}{a}\right)^2+t^2}}\)
\(=\frac{1}{a^2}.\frac{1}{\frac{b}{a}}\tan^{-1}{\left(\frac{t}{\frac{b}{a}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\frac{x}{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a^2}.\frac{a}{b}\tan^{-1}{\left(\frac{at}{b}\right)}+c\)
\(=\frac{1}{ab}\tan^{-1}{\left(\frac{a\tan{x}}{b}\right)}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{dx}{a^2\cos^2{x}-b^2\sin^2{x}}}\)
\(=\int{\frac{dx}{b^2\cos^2{x}\left(\frac{a^2}{b^2}-\frac{\sin^2{x}}{\cos^2{x}}\right)}}\)
\(=\frac{1}{b^2}\int{\frac{\sec^2{x}dx}{\frac{a^2}{b^2}-\tan^2{x}}}\)
\(=\frac{1}{b^2}\int{\frac{\sec^2{x}dx}{\left(\frac{a}{b}\right)^2-\tan^2{x}}}\)
\(=\frac{1}{b^2}\int{\frac{dt}{\left(\frac{a}{b}\right)^2-t^2}}\)
\(=\frac{1}{b^2}.\frac{1}{2.\frac{a}{b}}\ln{\left|\frac{\frac{a}{b}+t}{\frac{a}{b}-t}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{b^2}.\frac{b}{2a}\ln{\left|\frac{a+bt}{a-bt}\right|}+c\) ➜Note লব ও হরের সহিত \(b\) গুণ করে।
\(=\frac{1}{2ab}\ln{\left|\frac{a+b\tan{x}}{a-b\tan{x}}\right|}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{dx}{25\sin^2{x}-16\cos^2{x}}}\)
\(=\int{\frac{dx}{25\cos^2{x}\left(\frac{\sin^2{x}}{\cos^2{x}}-\frac{16}{25}\right)}}\)
\(=\frac{1}{25}\int{\frac{\sec^2{x}dx}{\frac{\sin^2{x}}{\cos^2{x}}-\frac{16}{25}}}\)
\(=\frac{1}{25}\int{\frac{\sec^2{x}dx}{\tan^2{x}-\frac{16}{25}}}\)
\(=\frac{1}{25}\int{\frac{dt}{t^2-\left(\frac{4}{5}\right)^2}}\)
\(=\frac{1}{25}.\frac{1}{2.\frac{4}{5}}\ln{\left|\frac{t-\frac{4}{5}}{t+\frac{4}{5}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{25}.\frac{5}{8}\ln{\left|\frac{5t-4}{5t+4}\right|}+c\) ➜Note লব ও হরের সহিত \(5\) গুণ করে।
\(=\frac{1}{5}.\frac{1}{8}\ln{\left|\frac{5t-4}{5t+4}\right|}+c\)
\(=\frac{1}{40}\ln{\left|\frac{5\tan{x}-4}{5\tan{x}+4}\right|}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{dx}{a^2+b^2\sin^2{x}}}\)
\(=\int{\frac{dx}{\cos^2{x}\left(a^2\frac{1}{\cos^2{x}}+b^2\frac{\sin^2{x}}{\cos^2{x}}\right)}}\)
\(=\int{\frac{\sec^2{x}dx}{a^2\sec^2{x}+b^2\tan^2{x}}}\)
\(=\int{\frac{\sec^2{x}dx}{a^2(1+\tan^2{x})+b^2\tan^2{x}}}\) ➜Note\(\because \sec^2{A}=1+\tan^2{A}\)
\(=\int{\frac{\sec^2{x}dx}{a^2+a^2\tan^2{x}+b^2\tan^2{x}}}\)
\(=\int{\frac{\sec^2{x}dx}{a^2+(a^2+b^2)\tan^2{x}}}\)
\(=\int{\frac{\sec^2{x}dx}{(a^2+b^2)\left(\frac{a^2}{a^2+b^2}+\tan^2{x}\right)}}\)
\(=\frac{1}{a^2+b^2}\int{\frac{\sec^2{x}dx}{\frac{a^2}{a^2+b^2}+\tan^2{x}}}\)
\(=\frac{1}{a^2+b^2}\int{\frac{dt}{\frac{a^2}{a^2+b^2}+t^2}}\)
\(=\frac{1}{a^2+b^2}\int{\frac{dt}{\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+t^2}}\)
\(=\frac{1}{a^2+b^2}.\frac{1}{\frac{a}{\sqrt{a^2+b^2}}}\tan^{-1}{\left(\frac{t}{\frac{a}{\sqrt{a^2+b^2}}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a^2+b^2}.\frac{\sqrt{a^2+b^2}}{a}\tan^{-1}{\left(\frac{t\sqrt{a^2+b^2}}{a}\right)}+c\)
\(=\frac{1}{\sqrt{a^2+b^2}\sqrt{a^2+b^2}}.\frac{\sqrt{a^2+b^2}}{a}\tan^{-1}{\left(\frac{\sqrt{a^2+b^2}}{a}t\right)}+c\)
\(=\frac{1}{a\sqrt{a^2+b^2}}\tan^{-1}{\left(\frac{\sqrt{a^2+b^2}}{a}\tan{x}\right)}+c\) ➜Note\(\because t=\tan{x}\)

\(\int{\frac{e^x}{1+e^{2x}}dx}\)
\(=\int{\frac{1}{1+(e^{x})^2}e^xdx}\)
\(=\int{\frac{1}{1+t^2}dt}\)
\(=\tan^{-1}{(t)}+c\) ➜Note \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan^{-1}{\left(e^x\right)}+c\) ➜Note\(\because t=e^x\)

\(\int{\frac{\sin{8x}}{9+\sin^4{4x}}dx}\)
\(=\int{\frac{1}{9+(\sin^2{4x})^2}\sin{8x}dx}\)
\(=\int{\frac{1}{9+t^2}.\frac{1}{4}dt}\)
\(=\frac{1}{4}\int{\frac{1}{3^2+t^2}dt}\)
\(=\frac{1}{4}.\frac{1}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{12}\tan^{-1}{\left(\frac{\sin^2{4x}}{3}\right)}+c\) ➜Note\(\because t=\sin^2{4x}\)
\(=\frac{1}{12}\tan^{-1}{\left(\frac{1}{3}\sin^2{4x}\right)}+c\)

\(\int{\frac{\sec^2{x}}{4+9\tan^2{x}}dx}\)
\(=\int{\frac{1}{4+9\tan^2{x}}\sec^2{x}dx}\)
\(=\int{\frac{1}{4+9t^2}dt}\)
\(=\int{\frac{1}{9\left(\frac{4}{9}+t^2\right)}dt}\)
\(=\frac{1}{9}\int{\frac{1}{\left(\frac{2}{3}\right)^2+t^2}dt}\)
\(=\frac{1}{9}.\frac{1}{\frac{2}{3}}\tan^{-1}{\left(\frac{t}{\frac{2}{3}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{9}.\frac{3}{2}\tan^{-1}{\left(\frac{3t}{2}\right)}+c\)
\(=\frac{1}{3}.\frac{1}{2}\tan^{-1}{\left(\frac{3\tan{x}}{2}\right)}+c\) ➜Note\(\because t=\tan{x}\)
\(=\frac{1}{6}\tan^{-1}{\left(\frac{3}{2}\tan{x}\right)}+c\)

\(\int{\frac{\cos{x}}{9+\sin^2{x}}dx}\)
\(=\int{\frac{1}{9+\sin^2{x}}\cos{x}dx}\)
\(=\int{\frac{1}{9+t^2}dt}\)
\(=\int{\frac{1}{3^2+t^2}dt}\)
\(=\frac{1}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan^{-1}{\left(\frac{\sin{x}}{3}\right)}+c\) ➜Note\(\because t=\sin{x}\)
\(=\frac{1}{3}\tan^{-1}{\left(\frac{1}{3}\sin{x}\right)}+c\)

\(\int{\frac{dx}{(1+x^2)\sqrt{1-(\tan^{-1}{x})^2}}}\)
\(=\int{\frac{1}{\sqrt{1-(\tan^{-1}{x})^2}}\times{\frac{dx}{(1+x^2)}}}\)
\(=\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{(\tan^{-1}{x})}+c\) ➜Note\(\because t=\tan^{-1}{x}\)

\(\int{\frac{x+35}{x^2-25}dx}\)
\(=\int{\left(\frac{x}{x^2-25}+\frac{35}{x^2-25}\right)dx}\)
\(=\int{\frac{x}{x^2-25}dx}+\int{\frac{35}{x^2-25}dx}\)
\(=\int{\frac{1}{x^2-25}xdx}+35\int{\frac{1}{x^2-5^2}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{2}dt}+35\int{\frac{1}{x^2-5^2}dx}\)
\(=\frac{1}{2}\int{\frac{1}{t}dt}+35\int{\frac{1}{x^2-5^2}dx}\)
\(=\frac{1}{2}\ln{\left|t\right|}+35.\frac{1}{2.5}\ln{\left|\frac{x-5}{x+5}\right|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{\left|x\right|}, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\ln{\left|x^2-25\right|}+7.\frac{1}{2}\ln{\left|\frac{x-5}{x+5}\right|}+c\) ➜Note\(\because t=x^2-25\)
\(=\frac{1}{2}\ln{\left|x^2-25\right|}+\frac{7}{2}\ln{\left|\frac{x-5}{x+5}\right|}+c\)

\(\int{\frac{x^2}{x^2-4}dx}\)
\(=\int{\frac{x^2-4+4}{x^2-4}dx}\)
\(=\int{\left(\frac{x^2-4}{x^2-4}+\frac{4}{x^2-4}\right)dx}\)
\(=\int{\frac{x^2-4}{x^2-4}dx}+\int{\frac{4}{x^2-4}dx}\)
\(=\int{dx}+4\int{\frac{1}{x^2-2^2}dx}\)
\(=x+4.\frac{1}{2.2}\ln{\left|\frac{x-2}{x+2}\right|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+4.\frac{1}{4}\ln{\left|\frac{x-2}{x+2}\right|}+c\)
\(=x+\ln{\left|\frac{x-2}{x+2}\right|}+c\)

\(\int{\frac{dx}{x\sqrt{x^4-1}}}\)
\(=\int{\frac{1}{\sqrt{(x^2)^2-1}}\times{\frac{dx}{x}}}\)
\(=\int{\frac{1}{\sqrt{t^2-1}}\times{\frac{1}{2t}dt}}\)
\(=\frac{1}{2}\int{\frac{1}{t\sqrt{t^2-1}}dt}\)
\(=\frac{1}{2}\sec^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{x\sqrt{x^2-1}}dx}=\sec^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sec^{-1}{(x^2)}+c\) ➜Note\(\because t=x^2\)

\(\int{\frac{x^2-1}{x^2-4}dx}\)
\(=\int{\frac{x^2-4+3}{x^2-4}dx}\)
\(=\int{\left(\frac{x^2-4}{x^2-4}+\frac{3}{x^2-4}\right)dx}\)
\(=\int{\frac{x^2-4}{x^2-4}dx}+\int{\frac{3}{x^2-4}dx}\)
\(=\int{dx}+3\int{\frac{1}{x^2-2^2}dx}\)
\(=x+3.\frac{1}{2.2}\ln{\left|\frac{x-2}{x+2}\right|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{3}{4}\ln{\left|\frac{x-2}{x+2}\right|}+c\)

\(\int{\frac{dx}{(a^2-x^2)^{\frac{3}{2}}}}\)
\(=\int{\frac{1}{(a^2-x^2)^{\frac{3}{2}}}dx}\)
\(=\int{\frac{1}{(a^2-a^2\sin^2{\theta})^{\frac{3}{2}}}a\cos{\theta}d\theta}\)
\(=\int{\frac{1}{\{a^2(1-\sin^2{\theta})\}^{\frac{3}{2}}}acos{\theta}d\theta}\)
\(=\int{\frac{1}{\{a^2\cos^2{\theta}\}^{\frac{3}{2}}}a\cos{\theta}d\theta}\) ➜Note\(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\frac{1}{a^3\cos^3{\theta}}a\cos{\theta}d\theta}\)
\(=\frac{1}{a^2}\int{\frac{1}{\cos^2{\theta}}d\theta}\)
\(=\frac{1}{a^2}\int{\sec^2{\theta}d\theta}\)
\(=\frac{1}{a^2}\tan{\theta}+c\) ➜Note \(\because \int{\sec^2{x}dx}=\tan{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a^2}\tan{\left\{\sin^{-1}{\left(\frac{x}{a}\right)}\right\}}+c\) ➜Note\(\because \theta=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{a^2}\tan{\left\{\tan^{-1}{\left(\frac{x}{\sqrt{a^2-x^2}}\right)}\right\}}+c\) ➜Note\(\because \sin^{-1}{\left(\frac{A}{B}\right)}=\tan^{-1}{\frac{A}{\sqrt{B^2-A^2}}}\)
\(=\frac{1}{a^2}\times{\frac{x}{\sqrt{a^2-x^2}}}+c\)
\(=\frac{x}{a^2\sqrt{a^2-x^2}}+c\)

\(\int{\frac{x^2dx}{x^{4}+a^{4}}}\)
\(=\int{\frac{2x^2}{2(x^{4}+a^{4})}dx}\)
\(=\int{\frac{(x^2+a^2)+(x^2-a^2)}{2(x^{4}+a^{4})}dx}\)
\(=\int{\left\{\frac{x^2+a^2}{2(x^{4}+a^{4})}+\frac{x^2-a^2}{2(x^{4}+a^{4})}\right\}dx}\)
\(=\frac{1}{2}\int{\frac{x^2+a^2}{x^{4}+a^{4}}dx}+\frac{1}{2}\int{\frac{x^2-a^2}{x^{4}+a^{4}}dx}\)
\(=\frac{1}{2}\int{\frac{1+\frac{a^2}{x^2}}{x^{2}+\frac{a^{4}}{x^2}}dx}+\frac{1}{2}\int{\frac{1-\frac{a^2}{x^2}}{x^{2}+\frac{a^{4}}{x^2}}dx}\) ➜Note লব ও হরের সহিত \(x^2\) ভাগ করে।
\(=\frac{1}{2}\int{\frac{1+\frac{a^2}{x^2}}{\left(x-\frac{a^{2}}{x}\right)^2+2.x.\frac{a^{2}}{x}}dx}\)\(+\frac{1}{2}\int{\frac{1-\frac{a^2}{x^2}}{\left(x+\frac{a^{2}}{x}\right)^2-2.x.\frac{a^{2}}{x}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\left(x-\frac{a^{2}}{x}\right)^2+2a^{2}}\left(1+\frac{a^2}{x^2}\right)dx}\)\(+\frac{1}{2}\int{\frac{1}{\left(x+\frac{a^{2}}{x}\right)^2-2a^{2}}\left(1-\frac{a^2}{x^2}\right)dx}\)
\(=\frac{1}{2}\int{\frac{1}{t^2_{1}+(\sqrt{2}a)^2}dt_{1}}+\frac{1}{2}\int{\frac{1}{t^2_{2}-(\sqrt{2}a)^2}dt_{2}}\)
\(=\frac{1}{2}\int{\frac{1}{(\sqrt{2}a)^2+t^2_{1}}dt_{1}}+\frac{1}{2}\int{\frac{1}{t^2_{2}-(\sqrt{2}a)^2}dt_{2}}\)
\(=\frac{1}{2}.\frac{1}{\sqrt{2}a}\tan^{-1}{\left(\frac{t_{1}}{\sqrt{2}a}\right)}+\frac{1}{2}.\frac{1}{2.\sqrt{2}a}\ln{\left|\frac{t_{2}-\sqrt{2}a}{t_{2}+\sqrt{2}a}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\tan^{-1}{\left(\frac{x}{a}\right)}\), \(\int{\frac{1}{x^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2\sqrt{2}a}\tan^{-1}{\left(\frac{x-\frac{a^{2}}{x}}{\sqrt{2}a}\right)}+\frac{1}{4\sqrt{2}a}\ln{\left|\frac{x+\frac{a^{2}}{x}-\sqrt{2}a}{x+\frac{a^{2}}{x}+\sqrt{2}a}\right|}+c\) ➜Note\(\because t_{1}=x-\frac{a^{2}}{x}, t_{2}=x+\frac{a^{2}}{x}\)
\(=\frac{1}{2\sqrt{2}a}\tan^{-1}{\left(\frac{x^2-a^{2}}{\sqrt{2}ax}\right)}+\frac{1}{4\sqrt{2}a}\ln{\left|\frac{x^2+a^{2}-\sqrt{2}ax}{x^2+a^{2}+\sqrt{2}ax}\right|}+c\) ➜Note লব ও হরের সহিত \(x\) গুণ করে।
\(=\frac{1}{2\sqrt{2}a}\left\{\tan^{-1}{\left(\frac{x^2-a^{2}}{\sqrt{2}ax}\right)}+\frac{1}{2}\ln{\left|\frac{x^2+a^{2}-\sqrt{2}ax}{x^2+a^{2}+\sqrt{2}ax}\right|}\right\}+c\)

\(\int{\frac{dx}{(e^x-1)^2}}\)
\(=\int{\frac{1}{(e^x-1)^2}dx}\)
\(=\int{\frac{1}{t^2}\times{\frac{1}{t+1}dt}}\)
\(=\int{\frac{1}{t^2(t+1)}dt}\)
\(=\int{\frac{t^2-t(t+1)+(t+1)}{t^2(t+1)}dt}\)
\(=\int{\left\{\frac{t^2}{t^2(t+1)}-\frac{t(t+1)}{t^2(t+1)}+\frac{(t+1)}{t^2(t+1)}\right\}dt}\)
\(=\int{\left\{\frac{1}{t+1}-\frac{1}{t}+\frac{1}{t^2}\right\}dt}\)
\(=\int{\frac{1}{t+1}dt}-\int{\frac{1}{t}dt}+\int{\frac{1}{t^2}dt}\)
\(=\ln{|t+1|}-\ln{|t|}-\frac{1}{t}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}, \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\left|e^x-1+1\right|}-\ln{|e^x-1|}-\frac{1}{e^x-1}+c\) ➜Note\(\because t=e^x-1\)
\(=\ln{(e^x)}-\ln{|e^x-1|}-\frac{1}{e^x-1}+c\)
\(=\ln{\left|\frac{e^x}{e^x-1}\right|}-\frac{1}{e^x-1}+c\)

\(\int{\frac{dx}{(x+2)\sqrt{x+3}}}\)
\(=\int{\frac{1}{(x+2)\sqrt{x+3}}dx}\)
\(=\int{\frac{1}{(t^2-3+2)t}2tdt}\) ➜Note \(\because x=t^2-3\)
\(=2\int{\frac{1}{t^2-1}dt}\)
\(=2\int{\frac{1}{t^2-1^2}dt}\)
\(=2.\frac{1}{2.1}\ln{\left|\frac{t-1}{t+1}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\left|\frac{t-1}{t+1}\right|}+c\)
\(=\ln{\left|\frac{\sqrt{x+3}-1}{\sqrt{x+3}+1}\right|}+c\) ➜Note \(\because t=\sqrt{x+3}\)

\(\int{\frac{dx}{(2x+1)\sqrt{4x+3}}}\)
\(=\int{\frac{1}{(2x+1)\sqrt{4x+3}}dx}\)
\(=\int{\frac{1}{\left\{2\left(\frac{t^2}{4}-\frac{3}{4}\right)+1\right\}t}\frac{t}{2}dt}\) ➜Note \(\because x=\frac{t^2}{4}-\frac{3}{4}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{t^2}{2}-\frac{3}{2}+1}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{t^2-3+2}{2}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\frac{t^2-1}{2}}dt}\)
\(=\frac{1}{2}\int{\frac{2}{t^2-1}dt}\)
\(=\int{\frac{1}{t^2-1^2}dt}\)
\(=\frac{1}{2.1}\ln{\left|\frac{t-1}{t+1}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\ln{\left|\frac{t-1}{t+1}\right|}+c\)
\(=\frac{1}{2}\ln{\left|\frac{\sqrt{4x+3}-1}{\sqrt{4x+3}+1}\right|}+c\) ➜Note \(\because t=\sqrt{4x+3}\)

\(\int{\frac{dx}{(2x+1)\sqrt{x+3}}}\)
\(=\int{\frac{1}{(2x+1)\sqrt{x+3}}dx}\)
\(=\int{\frac{1}{\{2(t^2-3)+1\}t}2tdt}\) ➜Note \(\because x=t^2-3\)
\(=2\int{\frac{1}{2(t^2-3)+1}dt}\)
\(=2\int{\frac{1}{2t^2-6+1}dt}\)
\(=2\int{\frac{1}{2t^2-5}dt}\)
\(=2\int{\frac{1}{2\left(t^2-\frac{5}{2}\right)}dt}\)
\(=\int{\frac{1}{t^2-\frac{5}{2}}dt}\)
\(=\int{\frac{1}{t^2-\left(\sqrt{\frac{5}{2}}\right)^2}dt}\)
\(=\frac{1}{2.\sqrt{\frac{5}{2}}}\ln{\left|\frac{t-\sqrt{\frac{5}{2}}}{t+\sqrt{\frac{5}{2}}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}.\sqrt{2}.\frac{\sqrt{5}}{\sqrt{2}}}\ln{\left|\frac{t-\sqrt{\frac{5}{2}}}{t+\sqrt{\frac{5}{2}}}\right|}+c\)
\(=\frac{1}{\sqrt{10}}\ln{\left|\frac{\sqrt{x+3}-\frac{\sqrt{5}}{\sqrt{2}}}{\sqrt{x+3}+\frac{\sqrt{5}}{\sqrt{2}}}\right|}+c\) ➜Note \(\because t=\sqrt{x+3}\)
\(=\frac{1}{\sqrt{10}}\ln{\left|\frac{\sqrt{2}\sqrt{x+3}-\sqrt{5}}{\sqrt{2}\sqrt{x+3}+\sqrt{5}}\right|}+c\) ➜Note লব ও হরের সহিত \(\sqrt{2}\) গুণ করে।
\(=\frac{1}{\sqrt{10}}\ln{\left|\frac{\sqrt{2x+6}-\sqrt{5}}{\sqrt{2x+6}+\sqrt{5}}\right|}+c\)

\(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\)
\(=\int{\frac{dx}{(x^{\frac{1}{4}})^2-x^{\frac{1}{4}}}}\)
\(=\int{\frac{4t^3dt}{t^2-t}}\)
\(=4\int{\frac{t^3dt}{t(t-1)}}\)
\(=4\int{\frac{t^2}{t-1}dt}\)
\(=4\int{\frac{t^2-1+1}{t-1}dt}\)
\(=4\int{\frac{(t+1)(t-1)+1}{t-1}dt}\)
\(=4\int{\left\{\frac{(t+1)(t-1)}{t-1}+\frac{1}{t-1}\right\}dt}\)
\(=4\int{\left\{t+1+\frac{1}{t-1}\right\}dt}\)
\(=4\int{tdt}+4\int{dt}+4\int{\frac{1}{t-1}dt}\)
\(=4\frac{t^2}{2}+4t+4\ln{|t-1|}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2t^2+4t+4\ln{|t-1|}+c\)
\(=2(x^{\frac{1}{4}})^2+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\) ➜Note \(\because t=x^{\frac{1}{4}}\)
\(=2x^{\frac{1}{2}}+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\)
\(=2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)

\(\int{\frac{\sqrt{x}dx}{1+\sqrt[3]{x}}}\)
\(=\int{\frac{\sqrt{t^6}.6t^5dt}{1+\sqrt[3]{t^6}}}\)
\(=6\int{\frac{t^{\frac{6}{2}}t^5dt}{1+t^{\frac{6}{3}}}}\)
\(=6\int{\frac{t^{3}t^5dt}{1+t^{2}}}\)
\(=6\int{\frac{t^8dt}{1+t^{2}}}\)
\(=6\int{\frac{t^8}{1+t^{2}}dt}\)
\(=6\int{\frac{t^6(1+t^{2})-t^4(1+t^{2})+t^2(1+t^{2})-(1+t^{2})+1}{1+t^{2}}dt}\)
\(=6\int{\frac{t^6(1+t^{2})}{1+t^{2}}dt}-6\int{\frac{t^4(1+t^{2})}{1+t^2}dt}\)\(+6\int{\frac{t^2(1+t^{2})}{1+t^2}dt}\)\(-6\int{\frac{1+t^{2}}{1+t^2}dt}+6\int{\frac{1}{1+t^{2}}dt}\)
\(=6\int{t^6dt}-6\int{t^4dt}+6\int{t^2dt}-6\int{dt}+6\int{\frac{1}{1+t^{2}}dt}\)
\(=6\frac{t^{6+1}}{6+1}-6\frac{t^{4+1}}{4+1}+6\frac{t^{2+1}}{2+1}-6t+6\tan^{-1}{t}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=6\frac{t^{7}}{7}-6\frac{t^{5}}{5}+6\frac{t^{3}}{3}-6t+6\tan^{-1}{t}+c\)
\(=\frac{6}{7}t^{7}-\frac{6}{5}t^{5}+\frac{6}{3}t^{3}-6t+6\tan^{-1}{t}+c\)
\(=\frac{6}{7}t^{7}-\frac{6}{5}t^{5}+2t^{3}-6t+6\tan^{-1}{t}+c\)
\(=\frac{6}{7}\left(x^{\frac{1}{6}}\right)^{7}-\frac{6}{5}\left(x^{\frac{1}{6}}\right)^{5}+2\left(x^{\frac{1}{6}}\right)^{3}\)\(-6\left(x^{\frac{1}{6}}\right)+6\tan^{-1}{\left(x^{\frac{1}{6}}\right)}+c\) ➜Note \(\because t=x^{\frac{1}{6}}\)
\(=\frac{6}{7}x^{\frac{7}{6}}-\frac{6}{5}x^{\frac{5}{6}}+2x^{\frac{1}{2}}-6x^{\frac{1}{6}}\)\(+6\tan^{-1}{\left(x^{\frac{1}{6}}\right)}+c\)

\(\int{\frac{1}{1+\tan{x}}dx}\)
\(=\int{\frac{1}{1+\frac{\sin{x}}{\cos{x}}}dx}\)
\(=\int{\frac{\cos{x}}{\cos{x}+\sin{x}}dx}\) ➜Note লব ও হরের সহিত \(\cos{x}\) গুণ করে।
\(=\frac{1}{2}\int{\frac{2\cos{x}}{\cos{x}+\sin{x}}dx}\)
\(=\frac{1}{2}\int{\frac{(\cos{x}+\sin{x})+(\cos{x}-\sin{x})}{\cos{x}+\sin{x}}dx}\)
\(=\frac{1}{2}\int{\frac{\cos{x}+\sin{x}}{\cos{x}+\sin{x}}dx}+\frac{1}{2}\int{\frac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{\cos{x}+\sin{x}}(\cos{x}-\sin{x})dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\frac{1}{2}x+\frac{1}{2}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{2}\ln{|\cos{x}+\sin{x}|}+c\) ➜Note \(\because t=\cos{x}+\sin{x}\)
\(=\frac{1}{2}(x+\ln{|\cos{x}+\sin{x}|})+c\)

\(\int{\frac{dx}{2+\cot{x}}}\)
\(=\int{\frac{1}{2+\cot{x}}dx}\)
\(=\int{\frac{1}{2+\frac{\cos{x}}{\sin{x}}}dx}\)
\(=\int{\frac{\sin{x}}{2\sin{x}+\cos{x}}dx}\) ➜Note লব ও হরের সহিত \(\sin{x}\) গুণ করে।
\(=\frac{1}{5}\int{\frac{5\sin{x}}{2\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{5}\int{\frac{2(2\sin{x}+\cos{x})-(2\cos{x}-\sin{x})}{2\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{5}\int{\frac{2(2\sin{x}+\cos{x})}{2\sin{x}+\cos{x}}dx}-\frac{1}{5}\int{\frac{2\cos{x}-\sin{x}}{2\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{5}\int{2dx}-\frac{1}{5}\int{\frac{1}{2\sin{x}+\cos{x}}(2\cos{x}-\sin{x})dx}\)
\(=\frac{2}{5}\int{dx}-\frac{1}{5}\int{\frac{1}{t}dt}\)
\(=\frac{2}{5}x-\frac{1}{5}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}x-\frac{1}{5}\ln{|2\sin{x}+\cos{x}|}+c\) ➜Note \(\because t=2\sin{x}+\cos{x}\)
\(=\frac{1}{5}(2x-\ln{|2\sin{x}+\cos{x}|})+c\)

\(\int{\frac{3\sin{x}dx}{\cos{x}+\sin{x}}}\)
\(=3\int{\frac{\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(=\frac{3}{2}\int{\frac{2\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(=\frac{3}{2}\int{\frac{(\cos{x}+\sin{x})-(\cos{x}-\sin{x})}{\cos{x}+\sin{x}}dx}\)
\(=\frac{3}{2}\int{\frac{\cos{x}+\sin{x}}{\cos{x}+\sin{x}}dx}-\frac{3}{2}\int{\frac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(=\frac{3}{2}\int{dx}-\frac{3}{2}\int{\frac{1}{\cos{x}+\sin{x}}(\cos{x}-\sin{x})dx}\)
\(=\frac{3}{2}\int{dx}-\frac{3}{2}\int{\frac{1}{t}dt}\)
\(=\frac{3}{2}x-\frac{3}{2}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{2}x-\frac{3}{2}\ln{|\cos{x}+\sin{x}|}+c\) ➜Note \(\because t=\cos{x}+\sin{x}\)
\(=\frac{3}{2}(x-\ln{|\cos{x}+\sin{x}|})+c\)

\(\int{\frac{dx}{a+b\cos{x}}}\)
\(=\int{\frac{1}{a+b\cos{x}}dx}\)
\(=\int{\frac{1}{a+b\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{a(1+\tan^2{\frac{x}{2}})+b(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{a+a\tan^2{\frac{x}{2}}+b-b\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{a+b+(a-b)\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{a+b+(a-b)t^2}2dt}\)
\(=2\int{\frac{1}{(a-b)\left\{\frac{a+b}{a-b}+t^2\right\}}dt}\)
\(=\frac{2}{a-b}\int{\frac{1}{\frac{a+b}{a-b}+t^2}dt}\)
\(=\frac{2}{a-b}\int{\frac{1}{\left(\sqrt{\frac{a+b}{a-b}}\right)^2+t^2}dt}\)
\(=\frac{2}{a-b}.\frac{1}{\sqrt{\frac{a+b}{a-b}}}\tan^{-1}{\left(\frac{t}{\sqrt{\frac{a+b}{a-b}}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{\sqrt{a-b}.\sqrt{a-b}}.\frac{\sqrt{a-b}}{\sqrt{a+b}}\tan^{-1}{\left(\frac{\sqrt{a-b}t}{\sqrt{a+b}}\right)}+c\)
\(=\frac{2}{\sqrt{a-b}}.\frac{1}{\sqrt{a+b}}\tan^{-1}{\left(\frac{\sqrt{a-b}\tan{\frac{x}{2}}}{\sqrt{a+b}}\right)}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)
\(=\frac{2}{\sqrt{(a-b)(a+b)}}\tan^{-1}{\left(\frac{\sqrt{a-b}\tan{\frac{x}{2}}}{\sqrt{a+b}}\right)}+c\)
\(=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}{\left(\frac{\sqrt{a-b}\tan{\frac{x}{2}}}{\sqrt{a+b}}\right)}+c\)

\(\int{\frac{dx}{1+\sin{x}-\cos{x}}}\)
\(=\int{\frac{1}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\frac{1}{1+\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)-\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-1+\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{2t^2+2t}2dt}\)
\(=\int{\frac{1}{2(t^2+t)}2dt}\)
\(=\int{\frac{1}{t^2+t}dt}\)
\(=\int{\frac{1}{t^2+2.t.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\frac{1}{4}}dt}\)
\(=\int{\frac{1}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}dt}\)
\(=\frac{1}{2.\frac{1}{2}}\ln{\left|\frac{t+\frac{1}{2}-\frac{1}{2}}{t+\frac{1}{2}+\frac{1}{2}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{1}\ln{\left|\frac{t}{t+\frac{1+1}{2}}\right|}+c\)
\(=\ln{\left|\frac{t}{t+\frac{2}{2}}\right|}+c\)
\(=\ln{\left|\frac{t}{t+1}\right|}+c\)
\(=\ln{\left|\frac{\tan{\frac{x}{2}}}{\tan{\frac{x}{2}}+1}\right|}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)
\(=\ln{\left|\frac{\tan{\frac{x}{2}}}{1+\tan{\frac{x}{2}}}\right|}+c\)

\(\int{\frac{dx}{a\cos{x}+b\sin{x}}}\)
\(=\int{\frac{1}{r\sin{\alpha}\cos{x}+r\cos{\alpha}\sin{x}}dx}\) ➜Note \(\because a=r\sin{\alpha}, b=r\cos{\alpha}\)
\(=\int{\frac{1}{r(\sin{\alpha}\cos{x}+\cos{\alpha}\sin{x})}dx}\)
\(=\int{\frac{1}{r\sin{(\alpha+x)}}dx}\) ➜Note \(\because \sin{A}\cos{B}+\cos{A}\sin{B}=\sin{(A+B)}\)
\(=\frac{1}{r}\int{\frac{1}{\sin{(x+\alpha)}}dx}\)
\(=\frac{1}{r}\int{cosec \ {(x+\alpha)}dx}\) ➜Note \(\because \frac{1}{\sin{A}}=cosec \ {A}\)
\(=\frac{1}{r}\ln{\left|\tan{\left(\frac{x+\alpha}{2}\right)}\right|}+c\) ➜Note \(\because \int{cosec \ {x}dx}=\ln{\left|\tan{\left(\frac{x}{2}\right)}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{a^2+b^2}}\ln{\left|\tan{\frac{x+\tan^{-1}{\left(\frac{a}{b}\right)}}{2}}\right|}+c\) ➜Note \(\because \alpha=\tan^{-1}{\left(\frac{a}{b}\right)}, r=\sqrt{a^2+b^2}\)
\(=\frac{1}{\sqrt{a^2+b^2}}\ln{\left|\tan{\frac{1}{2}\left\{x+\tan^{-1}{\left(\frac{a}{b}\right)}\right\}}\right|}+c\)

\(\int{\sqrt{\frac{1-x}{1+x}}dx}\)
\(=\int{\frac{\sqrt{1-x}}{\sqrt{1+x}}dx}\)
\(=\int{\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1+x}.\sqrt{1-x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{1-x}\) গুণ করে।
\(=\int{\frac{1-x}{\sqrt{(1+x)(1-x)}}dx}\)
\(=\int{\frac{1-x}{\sqrt{1-x^2}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}-\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}-\int{\frac{1}{t}\times{-tdt}}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}+\int{dt}\)
\(=\sin^{-1}{x}+t+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{x}+\sqrt{1-x^2}+c\) ➜Note \(\because t=\sqrt{1-x^2}\)

\(\int{\sqrt{\frac{1+x}{1-x}}dx}\)
\(=\int{\frac{\sqrt{1+x}}{\sqrt{1-x}}dx}\)
\(=\int{\frac{\sqrt{1+x}.\sqrt{1+x}}{\sqrt{1+x}.\sqrt{1-x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{1+x}\) গুণ করে।
\(=\int{\frac{1+x}{\sqrt{(1+x)(1-x)}}dx}\)
\(=\int{\frac{1+x}{\sqrt{1-x^2}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}+\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}+\int{\frac{1}{t}\times{-tdt}}\)
\(=\int{\frac{1}{\sqrt{1-x^2}}dx}-\int{dt}\)
\(=\sin^{-1}{x}-t+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{x}-\sqrt{1-x^2}+c\) ➜Note \(\because t=\sqrt{1-x^2}\)

\(\int{\sqrt{\frac{5-x}{5+x}}dx}\)
\(=\int{\frac{\sqrt{5-x}}{\sqrt{5+x}}dx}\)
\(=\int{\frac{\sqrt{5-x}.\sqrt{5-x}}{\sqrt{5+x}.\sqrt{5-x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{5-x}\) গুণ করে।
\(=\int{\frac{5-x}{\sqrt{(5+x)(5-x)}}dx}\)
\(=\int{\frac{5-x}{\sqrt{5^2-x^2}}dx}\)
\(=\int{\left(\frac{5}{\sqrt{5^2-x^2}}-\frac{x}{\sqrt{25-x^2}}\right)dx}\)
\(=5\int{\frac{1}{\sqrt{5^2-x^2}}dx}-\int{\frac{x}{\sqrt{25-x^2}}dx}\)
\(=5\int{\frac{1}{\sqrt{5^2-x^2}}dx}-\int{\frac{1}{t}\times{-tdt}}\)
\(=\int{\frac{1}{\sqrt{5^2-x^2}}dx}+\int{dt}\)
\(=\sin^{-1}{\left(\frac{x}{5}\right)}+t+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin^{-1}{\left(\frac{x}{5}\right)}+\sqrt{25-x^2}+c\) ➜Note \(\because t=\sqrt{25-x^2}\)

\(\int{\frac{dx}{\sqrt{(x-2)(x-3)}}}\)
\(=\int{\frac{2}{t}dt}\)
\(=2\int{\frac{1}{t}dt}\)
\(=2\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{|\sqrt{x-2}+\sqrt{x-3}|}+c\) ➜Note \(\because t=\sqrt{25-x^2}\)

\(\int{\frac{dx}{\sqrt{(x-2)(3-x)}}}\)
\(=\int{\frac{1}{\sqrt{(x-2)}\sqrt{(3-x)}}dx}\)
\(=\int{\frac{1}{t\sqrt{3-(t^2+2)}}2tdt}\)
\(=2\int{\frac{1}{\sqrt{3-t^2-2}}dt}\)
\(=2\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=2\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sin^{-1}{(\sqrt{x-2})}+c\) ➜Note \(\because t=\sqrt{x-2}\)

\(\int{\sqrt{\frac{a+x}{x}}dx}\)
\(=\int{\frac{\sqrt{a+x}}{\sqrt{x}}dx}\)
\(=\int{\frac{\sqrt{a+x}.\sqrt{a+x}}{\sqrt{x}.\sqrt{a+x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{a+x}\) গুণ করে।
\(=\int{\frac{a+x}{\sqrt{x(a+x)}}dx}\)
\(=\frac{1}{2}\int{\frac{2a+2x}{\sqrt{ax+x^2}}dx}\)
\(=\frac{1}{2}\int{\frac{a+a+2x}{\sqrt{ax+x^2}}dx}\)
\(=\frac{1}{2}\int{\left(\frac{a}{\sqrt{ax+x^2}}+\frac{1}{2}\frac{a+2x}{\sqrt{ax+x^2}}\right)dx}\)
\(=\frac{1}{2}\int{\frac{a}{\sqrt{x^2+ax}}dx}+\frac{1}{2}\int{\frac{a+2x}{\sqrt{ax+x^2}}dx}\)
\(=\frac{a}{2}\int{\frac{1}{\sqrt{x^2+2.x.\frac{a}{2}+\left(\frac{a}{2}\right)^2\)\(-\left(\frac{a}{2}\right)^2}}dx}\)\(+\frac{1}{2}\int{\frac{1}{\sqrt{ax+x^2}}(a+2x)dx}\)
\(=\frac{a}{2}\int{\frac{1}{\sqrt{\left(x+\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2}}dx}+\frac{1}{2}\int{\frac{1}{t}.2tdt}\)
\(=\frac{a}{2}\int{\frac{1}{\sqrt{\left(x+\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2}}dx}+\frac{1}{2}.2\int{dt}\)
\(=\frac{a}{2}\int{\frac{1}{\sqrt{\left(x+\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2}}dx}+\int{dt}\)
\(=\frac{a}{2}\ln{\left|\sqrt{\left(x+\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2}+\left(x+\frac{a}{2}\right)\right|}+t+c\) ➜Note \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|\sqrt{x^2-a^2}+x|}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a}{2}\ln{\left|\sqrt{x^2+2.x.\frac{a}{2}+\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2}+\left(x+\frac{a}{2}\right)\right|}+\sqrt{ax+x^2}+c\) ➜Note \(\because t=\sqrt{ax+x^2}\)
\(=\frac{a}{2}\ln{\left|\sqrt{x^2+ax}+\left(x+\frac{a}{2}\right)\right|}+\sqrt{ax+x^2}+c\)
\(=\sqrt{ax+x^2}+\frac{a}{2}\ln{\left|\sqrt{x^2+ax}+\left(x+\frac{a}{2}\right)\right|}+c\)

\(\int{(\sqrt{\tan{y}}+\sqrt{\cot{y}})dy}\)
\(=\int{\left(\frac{\sqrt{\sin{y}}}{\sqrt{\cos{y}}}+\frac{\sqrt{\cos{y}}}{\sqrt{\sin{y}}}\right)dy}\)
\(=\int{\left(\frac{\sin{y}+\cos{y}}{\sqrt{\sin{y}\cos{y}}}\right)dy}\)
\(=\int{\left(\frac{\sqrt{2}(\sin{y}+\cos{y})}{\sqrt{2\sin{y}\cos{y}}}\right)dy}\)
\(=\sqrt{2}\int{\left(\frac{\sin{y}+\cos{y}}{\sqrt{1-1+2\sin{y}\cos{y}}}\right)dy}\)
\(=\sqrt{2}\int{\left(\frac{\sin{y}+\cos{y}}{\sqrt{1-(1-2\sin{y}\cos{y})}}\right)dy}\)
\(=\sqrt{2}\int{\left(\frac{\sin{y}+\cos{y}}{\sqrt{1-(\sin^2{y}+\cos^2{y}-2\sin{y}\cos{y})}}\right)dy}\)
\(=\sqrt{2}\int{\left(\frac{\sin{y}+\cos{y}}{\sqrt{1-(\sin{y}-\cos{y})^2}}\right)dy}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-(\sin{y}-\cos{y})^2}}(\sin{y}+\cos{y})dy}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=\sqrt{2}\sin^{-1}{t}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sqrt{2}\sin^{-1}{\sin{y}-\cos{y}}+c\) ➜Note\(\because t=\sin{y}-\cos{y}\)

\(\int{\sqrt{\tan{y}}dy}\)
\(=\int{t\times{\frac{2t}{1+t^4}}dt}\)
\(=\int{\frac{2t^2}{1+t^4}dt}\)
\(=\int{\frac{(t^2+1)+(t^2-1)}{1+t^4}dt}\)
\(=\int{\frac{t^2+1}{1+t^4}dt}+\int{\frac{t^2-1}{1+t^4}dt}\)
\(=\int{\frac{t^2+1}{t^4+1}dt}+\int{\frac{t^2-1}{t^4+1}dt}\)
\(=\int{\frac{t^2\left(1+\frac{1}{t^2}\right)}{t^2\left(t^2+\frac{1}{t^2}\right)}dt}\)\(+\int{\frac{t^2\left(1-\frac{1}{t^2}\right)}{t^2\left(t^2+\frac{1}{t^2}\right)}dt}\)
\(=\int{\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt}+\int{\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt}\)
\(=\int{\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2.t.\frac{1}{t}}dt}+\int{\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2.t.\frac{1}{t}}dt}\)
\(=\int{\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}dt}+\int{\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2}dt}\)
\(=\int{\frac{1}{\left(t-\frac{1}{t}\right)^2+2}\left(1+\frac{1}{t^2}\right)dt}\)\(+\int{\frac{1}{\left(t+\frac{1}{t}\right)^2-2}\left(1-\frac{1}{t^2}\right)dt}\)
\(=\int{\frac{1}{z^2_{1}+(\sqrt{2})^2}dz_{1}}+\int{\frac{1}{z^2_{2}-(\sqrt{2})^2}dz_{2}}\)
\(=\int{\frac{1}{(\sqrt{2})^2+z^2_{1}}dz_{1}}+\int{\frac{1}{z^2_{2}-(\sqrt{2})^2}dz_{2}}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{z_{1}}{\sqrt{2}}\right)}+\frac{1}{2\sqrt{2}}\ln{\left|\frac{z_{2}-\sqrt{2}}{z_{2}+\sqrt{2}}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\), \(\int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)}+\frac{1}{2\sqrt{2}}\ln{\left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|}+c\) ➜Note\(\because z_{1}=t-\frac{1}{t}, z_{2}=t+\frac{1}{t}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{t^2-1}{\sqrt{2}t}\right)}+\frac{1}{2\sqrt{2}}\ln{\left|\frac{t^2+1-\sqrt{2}t}{t^2+1+\sqrt{2}t}\right|}+c\) ➜Note লব ও হরের সহিত \(t\) গুণ করে।
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{(\sqrt{\tan{y}})^2-1}{\sqrt{2}\sqrt{\tan{y}}}\right)}+\frac{1}{2\sqrt{2}}\ln{\left|\frac{(\sqrt{\tan{y}})^2+1-\sqrt{2}\sqrt{\tan{y}}}{(\sqrt{\tan{y}})^2+1+\sqrt{2}\sqrt{\tan{y}}}\right|}+c\) ➜Note\(\because t=\sqrt{\tan{y}}\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{\tan{y}-1}{\sqrt{2}\sqrt{\tan{y}}}\right)}\)\(+\frac{1}{2\sqrt{2}}\ln{\left|\frac{\tan{y}+1-\sqrt{2}\sqrt{\tan{y}}}{\tan{y}+1+\sqrt{2}\sqrt{\tan{y}}}\right|}+c\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{\tan{y}-1}{\sqrt{2\tan{y}}}\right)}\)\(+\frac{1}{2\sqrt{2}}\ln{\left|\frac{\tan{y}+1-\sqrt{2\tan{y}}}{\tan{y}+1+\sqrt{2\tan{y}}}\right|}+c\)
\(=\frac{1}{\sqrt{2}}\tan^{-1}{\left(\frac{\tan{y}-1}{\sqrt{2\tan{y}}}\right)}\)\(+\frac{1}{2\sqrt{2}}\ln{\left|\frac{\tan{y}-\sqrt{2\tan{y}}+1}{\tan{y}+\sqrt{2\tan{y}}+1}\right|}+c\)

\(\int{\frac{\sin{y}dy}{\sin{(y+a)}}}\)
\(=\int{\frac{\sin{\{(y+a)-a\}}}{\sin{(y+a)}}dy}\)
\(=\int{\frac{\sin{(y+a)}\cos{a}-\cos{(y+a)}\sin{a}}{\sin{(y+a)}}dy}\) ➜Note \(\because \sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(=\int{\left\{\frac{\sin{(y+a)}\cos{a}}{\sin{(y+a)}}-\frac{\cos{(y+a)}\sin{a}}{\sin{(y+a)}}\right\}dy}\)
\(=\int{\left\{\cos{a}-\sin{a}\frac{\cos{(y+a)}}{\sin{(y+a)}}\right\}dy}\)
\(=\int{\cos{a}dy}-\sin{a}\int{\frac{\cos{(y+a)}}{\sin{(y+a)}}dy}\)
\(=\cos{a}\int{dy}-\sin{a}\int{\frac{1}{\sin{(y+a)}}\cos{(y+a)}dy}\)
\(=\cos{a}\int{dy}-\sin{a}\int{\frac{1}{t}dt}\)
\(=\cos{a}.y-\sin{a}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=y\cos{a}-\sin{a}\ln{|\sin{(y+a)}|}+c\) ➜Note \(\because t=\sin{(y+a)}\)

\(\int{\frac{dy}{\sqrt{y}+\sqrt{1-y}}}\)
\(=\int{\frac{1}{\sqrt{y}+\sqrt{1-y}}dy}\)
\(=\int{\frac{1}{\sin{\theta}+\sqrt{1-\sin^2{\theta}}}2\sin{\theta}\cos{\theta}d\theta}\) ➜Note \(\because \sqrt{y}=\sin{\theta}\Rightarrow y=\sin^2{\theta}\)
\(=\int{\frac{2\sin{\theta}\cos{\theta}}{\sin{\theta}+\cos{\theta}}d\theta}\) ➜Note \(\because \sqrt{1-\sin^2{A}}=\cos{A}\)
\(=\int{\frac{1+2\sin{\theta}\cos{\theta}-1}{\sin{\theta}+\cos{\theta}}d\theta}\)
\(=\int{\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}-1}{\sin{\theta}+\cos{\theta}}d\theta}\) ➜Note \(\because \sin^2{A}+\cos^2{A}=1\)
\(=\int{\frac{(\sin{\theta}+\cos{\theta})^2-1}{\sin{\theta}+\cos{\theta}}d\theta}\) ➜Note \(\because (a+b)^2=a^2+b^2+2ab\)
\(=\int{\left\{\frac{(\sin{\theta}+\cos{\theta})^2}{\sin{\theta}+\cos{\theta}}-\frac{1}{\sin{\theta}+\cos{\theta}}\right\}d\theta}\)
\(=\int{\frac{(\sin{\theta}+\cos{\theta})^2}{\sin{\theta}+\cos{\theta}}d\theta}-\int{\frac{1}{\sin{\theta}+\cos{\theta}}d\theta}\)
\(=\int{(\sin{\theta}+\cos{\theta})d\theta}-\int{\frac{1}{\sqrt{2}\left(\sin{\theta}\frac{1}{\sqrt{2}}+\cos{\theta}\frac{1}{\sqrt{2}}\right)}d\theta}\)
\(=\int{(\sin{\theta}+\cos{\theta})d\theta}-\frac{1}{\sqrt{2}}\int{\frac{1}{\sin{\theta}\cos{\frac{\pi}{4}}+\cos{\theta}\sin{\frac{\pi}{4}}}d\theta}\) ➜Note \(\because \sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\)
\(=\int{(\sin{\theta}+\cos{\theta})d\theta}-\frac{1}{\sqrt{2}}\int{\frac{1}{\sin{\left(\theta+\frac{\pi}{4}\right)}}d\theta}\)
\(=\int{(\sin{\theta}+\cos{\theta})d\theta}-\frac{1}{\sqrt{2}}\int{cosec \ {\left(\theta+\frac{\pi}{4}\right)}d\theta}\)
\(=-\cos{\theta}+\sin{\theta}-\frac{1}{\sqrt{2}}\ln{\left|\tan{\frac{\left(\theta+\frac{\pi}{4}\right)}{2}}\right|}+c\) ➜Note \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\), \(\int{cosec \ {x}dx}=\ln{|\tan{\frac{x}{2}}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin{\theta}-\cos{\theta}-\frac{1}{\sqrt{2}}\ln{\left|\tan{\frac{1}{2}\left(\theta+\frac{\pi}{4}\right)}\right|}+c\)
\(=\sin{\theta}-\sqrt{1-\sin^2{\theta}}-\frac{1}{\sqrt{2}}\ln{\left|\tan{\frac{1}{2}\left(\theta+\frac{\pi}{4}\right)}\right|}+c\)| Note \(\because \cos{A}=\sqrt{1-\sin^2{A}}\)
\(=\sqrt{y}-\sqrt{1-y}-\frac{1}{\sqrt{2}}\ln{\left|\tan{\frac{1}{2}\left\{\sin^{-1}{(\sqrt{y})}+\frac{\pi}{4}\right\}}\right|}+c\) ➜Note \(\because \sqrt{y}=\sin{\theta}, y=\sin^2{\theta}, \theta=\sin^{-1}{(\sqrt{y})}\)

\(\int{\frac{dx}{4+5\sin{x}}}\)
\(=\int{\frac{1}{4+5\sin{x}}dx}\)
\(=\int{\frac{1}{4+5\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{4(1+\tan^2{\frac{x}{2}})+5(2\tan{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{4+4\tan^2{\frac{x}{2}}+10\tan{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{\sec^2{\frac{x}{2}}}{4\tan^2{\frac{x}{2}}+10\tan{\frac{x}{2}}+4}dx}\)
\(=\int{\frac{1}{4\tan^2{\frac{x}{2}}+10\tan{\frac{x}{2}}+4}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{4t^2+10t+4}2dt}\)
\(=2\int{\frac{1}{4\left(t^2+\frac{5}{2}t+1\right)}dt}\)
\(=2.\frac{1}{4}\int{\frac{1}{t^2+\frac{5}{2}t+1}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t^2+\frac{5}{2}t+1}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t^2+2.t.\frac{5}{4}+\left(\frac{5}{4}\right)^2-\frac{25}{16}+1}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\left(t+\frac{5}{4}\right)^2+1-\frac{25}{16}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\left(t+\frac{5}{4}\right)^2+\frac{16-25}{16}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\left(t+\frac{5}{4}\right)^2+\frac{-9}{16}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\left(t+\frac{5}{4}\right)^2-\frac{9}{16}}dt}\)
\(=\frac{1}{2}\int{\frac{1}{\left(t+\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2}dt}\)
\(=\frac{1}{2}.\frac{1}{2.\frac{3}{4}}\ln{\left|\frac{t+\frac{5}{4}-\frac{3}{4}}{t+\frac{5}{4}+\frac{3}{4}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}.\frac{1}{\frac{3}{2}}\ln{\left|\frac{4t+5-3}{4t+5+3}\right|}+c\) ➜Note লব ও হরের সহিত \(4\) গুণ করে।
\(=\frac{1}{2}.\frac{2}{3}\ln{\left|\frac{4t+2}{4t+8}\right|}+c\)
\(=\frac{1}{3}\ln{\left|\frac{2t+1}{2t+4}\right|}+c\) ➜Note লব ও হরের সহিত \(2\) ভাগ করে।
\(=\frac{1}{3}\ln{\left|\frac{2\tan{\left(\frac{x}{2}\right)}+1}{2\tan{\left(\frac{x}{2}\right)}+4}\right|}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)

\(\int{\frac{dx}{5+4\cos{x}}}\)
\(=\int{\frac{1}{5+4\cos{x}}dx}\)
\(=\int{\frac{1}{5+4\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{5(1+\tan^2{\frac{x}{2}})+4(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{5+5\tan^2{\frac{x}{2}}+4-4\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{9+\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{9+t^2}2dt}\)
\(=2\int{\frac{1}{\left\{9+t^2\right\}}dt}\)
\(=2\int{\frac{1}{(3)^2+t^2}dt}\)
\(=2.\frac{1}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{3}\tan^{-1}{\left(\frac{t}{3}\right)}+c\)
\(=\frac{2}{3}\tan^{-1}{\left\{\frac{1}{3}\left(\tan{\frac{x}{2}}\right)\right\}}+c\) ➜Note \(\because t=\tan{\frac{x}{2}}\)

\(\int{\frac{dx}{3+2\cos{x}}}\)
\(=\int{\frac{1}{3+2\cos{x}}dx}\)
\(=\int{\frac{1}{3+2\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{3(1+\tan^2{\frac{x}{2}})+2(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{3+3\tan^2{\frac{x}{2}}+2-2\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{5+\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{5+t^2}2dt}\)
\(=2\int{\frac{1}{\left\{5+t^2\right\}}dt}\)
\(=2\int{\frac{1}{(\sqrt{5})^2+t^2}dt}\)
\(=2.\frac{1}{\sqrt{5}}\tan^{-1}{\left(\frac{t}{\sqrt{5}}\right)}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left(\frac{t}{\sqrt{5}}\right)}+c\)
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left(\frac{\tan{\frac{x}{2}}}{\sqrt{5}}\right)}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)
\(=\frac{2}{\sqrt{5}}\tan^{-1}{\left\{\frac{1}{\sqrt{5}}\tan{\left(\frac{x}{2}\right)}\right\}}+c\)

\(\int{\frac{dx}{4+5\cos{x}}}\)
\(=\int{\frac{1}{4+5\cos{x}}dx}\)
\(=\int{\frac{1}{4+5\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{4(1+\tan^2{\frac{x}{2}})+5(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{4+4\tan^2{\frac{x}{2}}+5-5\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{9-\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{9-t^2}2dt}\)
\(=2\int{\frac{1}{\left\{9-t^2\right\}}dt}\)
\(=2\int{\frac{1}{(3)^2-t^2}dt}\)
\(=2.\frac{1}{2.3}\ln{\left|\frac{3+t}{3-t}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left(\frac{a+x}{a-x}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\ln{\left|\frac{3+\tan{\left(\frac{x}{2}\right)}}{3-\tan{\left(\frac{x}{2}\right)}}\right|}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)

\(\int{\frac{dx}{5-13\sin{x}}}\)
\(=\int{\frac{1}{5-13\sin{x}}dx}\)
\(=\int{\frac{1}{5-13\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{5(1+\tan^2{\frac{x}{2}})-13(2\tan{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{5+5\tan^2{\frac{x}{2}}-26\tan{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{\sec^2{\frac{x}{2}}}{5\tan^2{\frac{x}{2}}-26\tan{\frac{x}{2}}+5}dx}\)
\(=\int{\frac{1}{5\tan^2{\frac{x}{2}}-26\tan{\frac{x}{2}}+5}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{5t^2-26t+5}2dt}\)
\(=2\int{\frac{1}{5\left(t^2-\frac{26}{5}t+1\right)}dt}\)
\(=2.\frac{1}{5}\int{\frac{1}{t^2-\frac{26}{5}t+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{t^2-\frac{26}{5}t+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{t^2-2.t.\frac{13}{5}+\left(\frac{13}{5}\right)^2-\frac{169}{25}+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t-\frac{13}{5}\right)^2+1-\frac{169}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t-\frac{13}{5}\right)^2+\frac{25-169}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t-\frac{13}{5}\right)^2+\frac{-144}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t-\frac{13}{5}\right)^2-\frac{144}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t-\frac{13}{5}\right)^2-\left(\frac{12}{5}\right)^2}dt}\)
\(=\frac{2}{5}.\frac{1}{2.\frac{12}{5}}\ln{\left|\frac{t-\frac{13}{5}-\frac{12}{5}}{t-\frac{13}{5}+\frac{12}{5}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}.\frac{5}{24}\ln{\left|\frac{5t-13-12}{5t-13+12}\right|}+c\) ➜Note লব ও হরের সহিত \(5\) গুণ করে।
\(=\frac{1}{12}\ln{\left|\frac{5t-25}{5t-1}\right|}+c\)
\(=\frac{1}{12}\ln{\left|\frac{5\tan{\left(\frac{x}{2}\right)}-25}{5\tan{\left(\frac{x}{2}\right)}-1}\right|}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)

\(\int{\frac{dx}{5+4\sin{x}}}\)
\(=\int{\frac{1}{5+4\sin{x}}dx}\)
\(=\int{\frac{1}{5+4\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{5(1+\tan^2{\frac{x}{2}})+4(2\tan{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{5+5\tan^2{\frac{x}{2}}+8\tan{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{\sec^2{\frac{x}{2}}}{5\tan^2{\frac{x}{2}}+8\tan{\frac{x}{2}}+5}dx}\)
\(=\int{\frac{1}{5\tan^2{\frac{x}{2}}+8\tan{\frac{x}{2}}+5}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{5t^2+8t+5}2dt}\)
\(=2\int{\frac{1}{5\left(t^2+\frac{8}{5}t+1\right)}dt}\)
\(=2.\frac{1}{5}\int{\frac{1}{t^2+\frac{8}{5}t+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{t^2+\frac{8}{5}t+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{t^2+2.t.\frac{4}{5}+\left(\frac{4}{5}\right)^2-\frac{16}{25}+1}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t+\frac{4}{5}\right)^2+1-\frac{16}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t+\frac{4}{5}\right)^2+\frac{25-16}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t+\frac{4}{5}\right)^2+\frac{9}{25}}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(t+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}dt}\)
\(=\frac{2}{5}\int{\frac{1}{\left(\frac{3}{5}\right)^2+\left(t+\frac{4}{5}\right)^2}dt}\)
\(=\frac{2}{5}.\frac{1}{\frac{3}{5}}\tan^{-1}{\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}.\frac{5}{3}\tan^{-1}{\left(\frac{5t+4}{3}\right|}+c\) ➜Note লব ও হরের সহিত \(5\) গুণ করে।
\(=\frac{2}{3}\tan^{-1}{\left(\frac{5\tan{\left(\frac{x}{2}\right)}+4}{3}\right|}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)
\(=\frac{2}{3}\tan^{-1}{\left[\frac{1}{3}\left\{5\tan{\left(\frac{x}{2}\right)}+4\right\}\right]}+c\)

\(\int{\frac{dx}{3+5\cos{x}}}\)
\(=\int{\frac{1}{3+5\cos{x}}dx}\)
\(=\int{\frac{1}{3+5\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\int{\frac{1+\tan^2{\frac{x}{2}}}{3(1+\tan^2{\frac{x}{2}})+5(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\int{\frac{\sec^2{\frac{x}{2}}}{3+3\tan^2{\frac{x}{2}}+5-5\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int{\frac{1}{8-2\tan^2{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=\int{\frac{1}{8-2t^2}2dt}\)
\(=2\int{\frac{1}{2(4-t^2)}dt}\)
\(=2.\frac{1}{2}\int{\frac{1}{4-t^2}dt}\)
\(=\int{\frac{1}{(2)^2-t^2}dt}\)
\(=\frac{1}{2.2}\ln{\left|\frac{2+t}{2-t}\right|}+c\) ➜Note \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left(\frac{a+x}{a-x}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\ln{\left|\frac{2+\tan{\left(\frac{x}{2}\right)}}{2-\tan{\left(\frac{x}{2}\right)}}\right|}+c\) ➜Note \(\because t=\tan{\left(\frac{x}{2}\right)}\)

ধরি,
\(5\sin{x}+4\cos{x}=L(2\sin{x}+3\cos{x})\)\(+M\left\{\frac{d}{dx}(2\sin{x}+3\cos{x})\right\}\)
\(\Rightarrow 5\sin{x}+4\cos{x}=L(2\sin{x}+3\cos{x})+M(2\cos{x}-3\sin{x}) ....(1)\)
\(=2L\sin{x}+3L\cos{x}+2M\cos{x}-3M\sin{x}\)
\(=(2L-3M)\sin{x}+(3L+2M)\cos{x}\)
\(\therefore 5\sin{x}+4\cos{x}=(2L-3M)\sin{x}+(3L+2M)\cos{x}\)
\(\Rightarrow (2L-3M)\sin{x}+(3L+2M)\cos{x}=5\sin{x}+4\cos{x} ....(2)\)
\(\Rightarrow 2L-3M=5 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 3L+2M=4.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\((3)\Rightarrow 2L-3M-5=0\)
\((4)\Rightarrow 3L+2M-4=0\)
\(\Rightarrow \frac{L}{12+10}=\frac{M}{-15+8}=\frac{1}{4+9}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{22}=\frac{M}{-7}=\frac{1}{13}\)
\(\Rightarrow \frac{L}{22}=\frac{1}{13}; \frac{M}{-7}=\frac{1}{13}\)
\(\therefore L=\frac{22}{13}; M=-\frac{7}{13}\)
\(L\) ও \(M\) এর মান \((1)\) এ বসিয়ে,
\(5\sin{x}+4\cos{x}=\frac{22}{13}(2\sin{x}+3\cos{x})-\frac{7}{13}(2\cos{x}-3\sin{x})\)
\(\int{\frac{5\sin{x}+4\cos{x}}{2\sin{x}+3\cos{x}}dx}\)
\(=\int{\frac{\frac{22}{13}(2\sin{x}+3\cos{x})-\frac{7}{13}(2\cos{x}-3\sin{x})}{2\sin{x}+3\cos{x}}dx}\)
\(=\int{\left\{\frac{\frac{22}{13}(2\sin{x}+3\cos{x})}{2\sin{x}+3\cos{x}}-\frac{\frac{7}{13}(2\cos{x}-3\sin{x})}{2\sin{x}+3\cos{x}}\right\}dx}\)
\(=\frac{22}{13}\int{dx}-\frac{7}{13}\int{\frac{2\cos{x}-3\sin{x}}{2\sin{x}+3\cos{x}}dx}\)
\(=\frac{22}{13}\int{dx}-\frac{7}{13}\int{\frac{1}{2\sin{x}+3\cos{x}}(2\cos{x}-3\sin{x})dx}\)
\(=\frac{22}{13}\int{dx}-\frac{7}{13}\int{\frac{1}{t}dt}\)
\(=\frac{22}{13}x-\frac{7}{13}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{22x}{13}-\frac{7}{13}\ln{\left|2\sin{x}+3\cos{x}\right|}+c\) ➜Note \(\because t=2\sin{x}+3\cos{x}\)

ধরি,
\(4\sin{x}+3\cos{x}=L(5\sin{x}+4\cos{x})\)\(+M\left\{\frac{d}{dx}(5\sin{x}+4\cos{x})\right\}\)
\(\Rightarrow 4\sin{x}+3\cos{x}=L(5\sin{x}+4\cos{x})+M(5\cos{x}-4\sin{x}) ....(1)\)
\(=5L\sin{x}+4L\cos{x}+5M\cos{x}-4M\sin{x}\)
\(=(5L-4M)\sin{x}+(4L+5M)\cos{x}\)
\(\therefore 4\sin{x}+3\cos{x}=(5L-4M)\sin{x}+(4L+5M)\cos{x}\)
\(\Rightarrow (5L-4M)\sin{x}+(4L+5M)\cos{x}=4\sin{x}+3\cos{x} ....(2)\)
\(\Rightarrow 5L-4M=4 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 4L+5M=3.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\((3)\Rightarrow 5L-4M-4=0\)
\((4)\Rightarrow 4L+5M-3=0\)
\(\Rightarrow \frac{L}{12+20}=\frac{M}{-16+15}=\frac{1}{25+16}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{32}=\frac{M}{-1}=\frac{1}{41}\)
\(\Rightarrow \frac{L}{32}=\frac{1}{41}; \frac{M}{-1}=\frac{1}{41}\)
\(\therefore L=\frac{32}{41}; M=-\frac{1}{41}\)
\(L\) ও \(M\) এর মান \((1)\) এ বসিয়ে,
\(4\sin{x}+3\cos{x}=\frac{32}{41}(5\sin{x}+4\cos{x})-\frac{1}{41}(5\cos{x}-4\sin{x})\)
\(\int{\frac{4\sin{x}+3\cos{x}}{5\sin{x}+4\cos{x}}dx}\)
\(=\int{\frac{\frac{32}{41}(5\sin{x}+4\cos{x})-\frac{1}{41}(5\cos{x}-4\sin{x})}{5\sin{x}+4\cos{x}}dx}\)
\(=\int{\left\{\frac{\frac{32}{41}(5\sin{x}+4\cos{x})}{5\sin{x}+4\cos{x}}-\frac{\frac{1}{41}(5\cos{x}-4\sin{x})}{5\sin{x}+4\cos{x}}\right\}dx}\)
\(=\frac{32}{41}\int{dx}-\frac{1}{41}\int{\frac{5\cos{x}-4\sin{x}}{5\sin{x}+4\cos{x}}dx}\)
\(=\frac{32}{41}\int{dx}-\frac{1}{41}\int{\frac{1}{5\sin{x}+4\cos{x}}(5\cos{x}-4\sin{x})dx}\)
\(=\frac{32}{41}\int{dx}-\frac{1}{41}\int{\frac{1}{t}dt}\)
\(=\frac{32}{41}x-\frac{1}{41}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{32x}{41}-\frac{1}{41}\ln{\left|5\sin{x}+4\cos{x}\right|}+c\) ➜Note \(\because t=5\sin{x}+4\cos{x}\)

ধরি,
\(11\cos{x}-16\sin{x}=L(2\cos{x}+5\sin{x})\)\(+M\left\{\frac{d}{dx}(2\cos{x}+5\sin{x})\right\}\)
\(\Rightarrow 11\cos{x}-16\sin{x}=L(2\cos{x}+5\sin{x})+M(-2\sin{x}+5\cos{x}) ....(1)\)
\(=2L\cos{x}+5L\sin{x}-2M\sin{x}+5M\cos{x}\)
\(=(2L+5M)\cos{x}+(5L-2M)\sin{x}\)
\(\therefore 11\cos{x}-16\sin{x}=(2L+5M)\cos{x}+(5L-2M)\sin{x}\)
\(\Rightarrow (2L+5M)\cos{x}+(5L-2M)\sin{x}=11\cos{x}-16\sin{x} ....(2)\)
\(\Rightarrow 5L-2M=-16 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 2L+5M=11.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\((3)\Rightarrow 5L-2M+16=0\)
\((4)\Rightarrow 2L+5M-11=0\)
\(\Rightarrow \frac{L}{22-80}=\frac{M}{32+55}=\frac{1}{25+4}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{-58}=\frac{M}{87}=\frac{1}{29}\)
\(\Rightarrow \frac{L}{-58}=\frac{1}{29}; \frac{M}{87}=\frac{1}{29}\)
\(\Rightarrow L=-\frac{58}{29}; M=\frac{87}{29}\)
\(\therefore L=-2; M=3\)
\(L\) ও \(M\) এর মান \((1)\) এ বসিয়ে,
\(11\cos{x}-16\sin{x}=-2(2\cos{x}+5\sin{x})+3(-2\sin{x}+5\cos{x})\)
\(\int{\frac{11\cos{x}-16\sin{x}}{2\cos{x}+5\sin{x}}dx}\)
\(=\int{\frac{-2(2\cos{x}+5\sin{x})+3(-2\sin{x}+5\cos{x})}{2\cos{x}+5\sin{x}}dx}\)
\(=\int{\left\{\frac{-2(2\cos{x}+5\sin{x})}{2\cos{x}+5\sin{x}}+\frac{3(-2\sin{x}+5\cos{x})}{2\cos{x}+5\sin{x}}\right\}dx}\)
\(=-2\int{dx}+3\int{\frac{-2\sin{x}+5\cos{x}}{2\cos{x}+5\sin{x}}dx}\)
\(=-2\int{dx}+3\int{\frac{1}{2\cos{x}+5\sin{x}}(-2\sin{x}+5\cos{x})dx}\)
\(=-2\int{dx}+3\int{\frac{1}{t}dt}\)
\(=-2\int{dx}+3\int{\frac{1}{t}dt}\)
\(=-2x+3\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=3\ln{|t|}-2x+c\)
\(=3\ln{|2\cos{x}+5\sin{x}|}-2x+c\) ➜Note \(\because t=2\cos{x}+5\sin{x}\)

ধরি,
\(2\sin{x}+3\cos{x}=L(3\sin{x}+4\cos{x})\)\(+M\left\{\frac{d}{dx}(3\sin{x}+4\cos{x})\right\}\)
\(\Rightarrow 2\sin{x}+3\cos{x}=L(3\sin{x}+4\cos{x})+M(3\cos{x}-4\sin{x}) ....(1)\)
\(=3L\sin{x}+4L\cos{x}+3M\cos{x}-4M\sin{x}\)
\(=(3L-4M)\sin{x}+(4L+3M)\cos{x}\)
\(\therefore 2\sin{x}+3\cos{x}=(3L-4M)\sin{x}+(4L+3M)\cos{x}\)
\(\Rightarrow (3L-4M)\sin{x}+(4L+3M)\cos{x}=2\sin{x}+3\cos{x} ....(2)\)
\(\Rightarrow 3L-4M=2 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 4L+3M=3.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\((3)\Rightarrow 3L-4M-2=0\)
\((4)\Rightarrow 4L+3M-3=0\)
\(\Rightarrow \frac{L}{12+6}=\frac{M}{-8+9}=\frac{1}{9+16}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{18}=\frac{M}{1}=\frac{1}{25}\)
\(\Rightarrow \frac{L}{18}=\frac{1}{25}; \frac{M}{1}=\frac{1}{25}\)
\(\therefore L=\frac{18}{25}; M=\frac{1}{25}\)
\(L\) ও \(M\) এর মান \((1)\) এ বসিয়ে,
\(2\sin{x}+3\cos{x}=\frac{18}{25}(3\sin{x}+4\cos{x})+\frac{1}{25}(3\cos{x}-4\sin{x})\)
\(\int{\frac{2\sin{x}+3\cos{x}}{3\sin{x}+4\cos{x}}dx}\)
\(=\int{\frac{\frac{18}{25}(3\sin{x}+4\cos{x})+\frac{1}{25}(3\cos{x}-4\sin{x})}{3\sin{x}+4\cos{x}}dx}\)
\(=\int{\left\{\frac{\frac{18}{25}(3\sin{x}+4\cos{x})}{3\sin{x}+4\cos{x}}+\frac{\frac{1}{25}(3\cos{x}-4\sin{x})}{3\sin{x}+4\cos{x}}\right\}dx}\)
\(=\frac{18}{25}\int{dx}+\frac{1}{25}\int{\frac{3\cos{x}-4\sin{x}}{3\sin{x}+4\cos{x}}dx}\)
\(=\frac{18}{25}\int{dx}+\frac{1}{25}\int{\frac{1}{3\sin{x}+4\cos{x}}(3\cos{x}-4\sin{x})dx}\)
\(=\frac{18}{25}\int{dx}+\frac{1}{25}\int{\frac{1}{t}dt}\)
\(=\frac{18}{25}x+\frac{1}{25}\ln{|t|}+c\)
\(=\frac{18}{25}x+\frac{1}{25}\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{18x}{25}+\frac{1}{25}\ln{|3\sin{x}+4\cos{x}|}+c\) ➜Note \(\because t=3\sin{x}+4\cos{x}\)

ধরি,
\(6+3\sin{x}+14\cos{x}=L(3+4\sin{x}+5\cos{x})\)\(+M\left\{\frac{d}{dx}(3+4\sin{x}+5\cos{x})\right\}+N\)
\(\Rightarrow 6+3\sin{x}+14\cos{x}=L(3+4\sin{x}+5\cos{x})+M(4\cos{x}-5\sin{x})+N ....(1)\)
\(=3L+4L\sin{x}+5L\cos{x})+4M\cos{x}-5M\sin{x}+N\)
\(=(4L-5M)\sin{x}+(5L+4M)\cos{x}+3L+N\)
\(\therefore 6+3\sin{x}+14\cos{x}=(4L-5M)\sin{x}+(5L+4M)\cos{x}+3L+N\)
\(\Rightarrow (4L-5M)\sin{x}+(5L+4M)\cos{x}+3L+N=3\sin{x}+14\cos{x}+6 ........(2)\)
\(\Rightarrow 4L-5M=3 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 5L+4M=14.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 3L+N=6.....(5)\) ➜Note \((2)\) এর উভয় পার্শ হতে ধ্রুবক রাশির সমতা নিয়ে ।
\((3)\Rightarrow 4L-5M-3=0\)
\((4)\Rightarrow 5L+4M-14=0\)
\(\Rightarrow \frac{L}{70+12}=\frac{M}{-15+56}=\frac{1}{16+25}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{82}=\frac{M}{41}=\frac{1}{41}\)
\(\Rightarrow \frac{L}{82}=\frac{1}{41}; \frac{M}{41}=\frac{1}{41}\)
\(\Rightarrow L=\frac{82}{41}; M=\frac{41}{41}\)
\(\therefore L=2; M=1\)
\(L\) এর মান \((5)\) এ বসিয়ে,
\(3.2+N=6\)
\(\Rightarrow 6+N=6\)
\(\Rightarrow N=6-6\)
\(\therefore N=0\)
\(L\), \(M\) ও \(N\) এর মান \((1)\) এ বসিয়ে,
\(6+3\sin{x}+14\cos{x}=2(3+4\sin{x}+5\cos{x})+1.(4\cos{x}-5\sin{x})+0\)
\(\therefore 6+3\sin{x}+14\cos{x}=2(3+4\sin{x}+5\cos{x})+(4\cos{x}-5\sin{x})\)
\(\int{\frac{6+3\sin{x}+14\cos{x}}{3+4\sin{x}+5\cos{x}}dx}\)
\(=\int{\frac{2(3+4\sin{x}+5\cos{x})+(4\cos{x}-5\sin{x})}{3+4\sin{x}+5\cos{x}}dx}\)
\(=\int{\left\{\frac{2(3+4\sin{x}+5\cos{x})}{3+4\sin{x}+5\cos{x}}+\frac{4\cos{x}-5\sin{x}}{3+4\sin{x}+5\cos{x}}\right\}dx}\)
\(=2\int{dx}+\int{\frac{4\cos{x}-5\sin{x}}{3+4\sin{x}+5\cos{x}}dx}\)
\(=2\int{dx}+\int{\frac{1}{3+4\sin{x}+5\cos{x}}(4\cos{x}-5\sin{x})dx}\)
\(=2\int{dx}+\int{\frac{1}{t}dt}\)
\(=2x+\ln{|t|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2x+\ln{|3+4\sin{x}+5\cos{x}|}+c\) ➜Note \(\because t=3+4\sin{x}+5\cos{x}\)

ধরি,
\(1-\sin{x}-\cos{x}=L(1+\sin{x}-\cos{x})\)\(+M\left\{\frac{d}{dx}(1+\sin{x}-\cos{x})\right\}+N\)
\(\Rightarrow 1-\sin{x}-\cos{x}=L(1+\sin{x}-\cos{x})+M(\cos{x}+\sin{x})+N ....(1)\)
\(=L+L\sin{x}-L\cos{x}+M\cos{x}+M\sin{x}+N\)
\(=(L+M)\sin{x}+(M-L)\cos{x}+L+N\)
\(\therefore 1-\sin{x}-\cos{x}=(L+M)\sin{x}+(M-L)\cos{x}+L+N\)
\(\Rightarrow (L+M)\sin{x}+(M-L)\cos{x}+L+N=-\sin{x}-\cos{x}+1 ........(2)\)
\(\Rightarrow L+M=-1 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow M-L=-1.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow L+N=1.....(5)\) ➜Note \((2)\) এর উভয় পার্শ হতে ধ্রুবক রাশির সমতা নিয়ে ।
\((3)+(4)\) এর সাহায্যে
\(L+M+M-L=-1-1\)
\(\Rightarrow 2M=-2\)
\(\therefore M=-1\)
\((3)\) হতে,
\(L-1=-1\)
\(\Rightarrow L=-1+1\)
\(\therefore L=0\)
\((5)\) হতে,
\(0+N=1\)
\(\therefore N=1\)
\(L\), \(M\) ও \(N\) এর মান \((1)\) এ বসিয়ে,
\(1-\sin{x}-\cos{x}=0.(1+\sin{x}-\cos{x})+(-1)(\cos{x}+\sin{x})+1\)
\(\Rightarrow 1-\sin{x}+\cos{x}=-(\cos{x}+\sin{x})+1\)
\(\int{\frac{1-\sin{x}+\cos{x}}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\frac{-(\cos{x}+\sin{x})+1}{1+\sin{x}-\cos{x}}dx}\)
\(=\int{\left\{\frac{-(\cos{x}+\sin{x})}{1+\sin{x}-\cos{x}}+\frac{1}{1+\sin{x}-\cos{x}}\right\}dx}\)
\(=\int{\frac{-(\cos{x}+\sin{x})}{1+\sin{x}-\cos{x}}dx}+\int{\frac{1}{1+\sin{x}-\cos{x}}dx}\)
\(=-\int{\frac{\cos{x}+\sin{x}}{1+\sin{x}-\cos{x}}dx}+\int{\frac{1}{1+\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)-\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=-\int{\frac{\cos{x}+\sin{x}}{1+\sin{x}-\cos{x}}dx}+\int{\frac{1+\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-(1-\tan^2{\frac{x}{2}})}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=-\int{\frac{\cos{x}+\sin{x}}{1+\sin{x}-\cos{x}}dx}+\int{\frac{\sec^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}-1+\tan^2{\frac{x}{2}}}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=-\int{\frac{1}{1+\sin{x}-\cos{x}}(\cos{x}+\sin{x})dx}+\int{\frac{1}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}}\sec^2{\frac{x}{2}}dx}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{2t^2_{2}+2t_{2}}2dt_{2}}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{2(t^2_{2}+t_{2})}2dt_{2}}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{t^2_{2}+t_{2}}dt_{2}}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{t^2_{2}+2.t_{2}.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}}dt_{2}}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{\left(t_{2}+\frac{1}{2}\right)^2-\frac{1}{4}}dt_{2}}\)
\(=-\int{\frac{1}{t_{1}}dt_{1}}+\int{\frac{1}{\left(t_{2}+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}dt_{2}}\)
\(=-\ln{|t_{1}|}+\frac{1}{2.\frac{1}{2}}\ln{\left|\frac{t_{2}+\frac{1}{2}-\frac{1}{2}}{t_{2}+\frac{1}{2}+\frac{1}{2}}\right|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{|t_{1}|}+\frac{1}{1}\ln{\left|\frac{t_{2}}{t_{2}+\frac{1+1}{2}}\right|}+c\)
\(=-\ln{|t_{1}|}+\ln{\left|\frac{t_{2}}{t_{2}+\frac{2}{2}}\right|}+c\)
\(=-\ln{|t_{1}|}+\ln{\left|\frac{t_{2}}{t_{2}+1}\right|}+c\)
\(=-\ln{|1+\sin{x}-\cos{x}|}+\ln{\left|\frac{\tan{\left(\frac{x}{2}\right)}}{\tan{\left(\frac{x}{2}\right)}+1}\right|}+c\) ➜Note \(\because t_{1}=1+\sin{x}-\cos{x}, t_{2}=\tan{\left(\frac{x}{2}\right)}\)
\(=\ln{\left|\frac{\tan{\left(\frac{x}{2}\right)}}{\tan{\left(\frac{x}{2}\right)}+1}\right|}-\ln{|1+\sin{x}-\cos{x}|}+c\)

ধরি,
\(\cos{x}+2\sin{x}+3=L(4\cos{x}+5\sin{x}+6)\)\(+M\left\{\frac{d}{dx}(4\cos{x}+5\sin{x}+6)\right\}+N\)
\(\Rightarrow \cos{x}+2\sin{x}+3=L(4\cos{x}+5\sin{x}+6)+M(-4\sin{x}+5\cos{x})+N\)
\(\Rightarrow \cos{x}+2\sin{x}+3=L(4\cos{x}+5\sin{x}+6)+M(5\cos{x}-4\sin{x})+N ....(1)\)
\(=4L\cos{x}+5L\sin{x}+6L+5M\cos{x}-4M\sin{x}+N\)
\(=(5L-4M)\sin{x}+(4L+5M)\cos{x}+6L+N\)
\(\therefore \cos{x}+2\sin{x}+3=(5L-4M)\sin{x}+(4L+5M)\cos{x}+6L+N\)
\(\Rightarrow (5L-4M)\sin{x}+(4L+5M)\cos{x}+6L+N=\cos{x}+2\sin{x}+3 ........(2)\)
\(\Rightarrow 5L-4M=2 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 4L+5M=1.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 6L+N=3.....(5)\) ➜Note \((2)\) এর উভয় পার্শ হতে ধ্রুবক রাশির সমতা নিয়ে ।
\((3)\Rightarrow 5L-4M-2=0\)
\((4)\Rightarrow 4L+5M-1=0\)
\(\Rightarrow \frac{L}{4+10}=\frac{M}{-8+5}=\frac{1}{25+16}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{14}=\frac{M}{-3}=\frac{1}{41}\)
\(\Rightarrow \frac{L}{14}=\frac{1}{41}; \frac{M}{-3}=\frac{1}{41}\)
\(\therefore L=\frac{14}{41}; M=-\frac{3}{41}\)
\(L\) এর মান \((5)\) এ বসিয়ে,
\(6.\frac{14}{41}+N=3\)
\(\Rightarrow \frac{84}{41}+N=3\)
\(\Rightarrow N=3-\frac{84}{41}\)
\(\Rightarrow N=\frac{123-84}{41}\)
\(\therefore N=\frac{39}{41}\)
\(L\), \(M\) ও \(N\) এর মান \((1)\) এ বসিয়ে,
\(\cos{x}+2\sin{x}+3=\frac{14}{41}(4\cos{x}+5\sin{x}+6)-\frac{3}{41}(5\cos{x}-4\sin{x})+\frac{39}{41}\) \(\int{\frac{\cos{x}+2\sin{x}+3}{4\cos{x}+5\sin{x}+6}dx}\)
\(=\int{\frac{\frac{14}{41}(4\cos{x}+5\sin{x}+6)-\frac{3}{41}(5\cos{x}-4\sin{x})+\frac{39}{41}}{4\cos{x}+5\sin{x}+6}dx}\)
\(=\int{\left\{\frac{\frac{14}{41}(4\cos{x}+5\sin{x}+6)}{4\cos{x}+5\sin{x}+6}\)\(-\frac{\frac{3}{41}(5\cos{x}-4\sin{x})}{4\cos{x}+5\sin{x}+6}+\frac{\frac{39}{41}}{4\cos{x}+5\sin{x}+6}\right\}dx}\)
\(=\frac{14}{41}\int{\frac{4\cos{x}+5\sin{x}+6}{4\cos{x}+5\sin{x}+6}dx}\)\(-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}+\frac{39}{41}\int{\frac{1}{4\cos{x}+5\sin{x}+6}dx}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}+\frac{39}{41}\int{\frac{1}{4\cos{x}+5\sin{x}+6}dx}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}\)
\(+\frac{39}{41}\int{\frac{1}{4\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)+5\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)+6}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}\)
\(+\frac{39}{41}\int{\frac{1+\tan^2{\frac{x}{2}}}{4\left(1-\tan^2{\frac{x}{2}}\right)+5\left(2\tan{\frac{x}{2}}\right)+6\left(1+\tan^2{\frac{x}{2}}\right)}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}\)\(+\frac{39}{41}\int{\frac{1+\tan^2{\frac{x}{2}}}{4-4\tan^2{\frac{x}{2}}+10\tan{\frac{x}{2}}+6+6\tan^2{\frac{x}{2}}}dx}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{5\cos{x}-4\sin{x}}{4\cos{x}+5\sin{x}+6}dx}\)\(+\frac{39}{41}\int{\frac{\sec^2{\left(\frac{x}{2}\right)}}{2\tan^2{\frac{x}{2}}+10\tan{\frac{x}{2}}+10}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{4\cos{x}+5\sin{x}+6}(5\cos{x}-4\sin{x})dx}\)\(+\frac{39}{41}\int{\frac{1}{2\left(\tan^2{\frac{x}{2}}+5\tan{\frac{x}{2}}+5\right)}\sec^2{\left(\frac{x}{2}\right)}dx}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{2(t^2_{2}+5t_{2}+5)}2dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{t^2_{2}+5t_{2}+5}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{t^2_{2}+2.t_{2}.\frac{5}{2}\)\(+\left(\frac{5}{2}\right)^2-\frac{25}{4}+5}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{\left(t_{2}+\frac{5}{2}\right)^2+5-\frac{25}{4}}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{\left(t_{2}+\frac{5}{2}\right)^2+\frac{20-25}{4}}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{\left(t_{2}+\frac{5}{2}\right)^2+\frac{-5}{4}}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{\left(t_{2}+\frac{5}{2}\right)^2-\frac{5}{4}}dt_{2}}\)
\(=\frac{14}{41}\int{dx}-\frac{3}{41}\int{\frac{1}{t_{1}}dt_{1}}+\frac{39}{41}\int{\frac{1}{\left(t_{2}+\frac{5}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2}dt_{2}}\)
\(=\frac{14}{41}x-\frac{3}{41}\ln{|t_{1}|}+\frac{39}{41}.\frac{1}{2.\frac{\sqrt{5}}{2}}\ln{\left|\frac{t_{2}+\frac{5}{2}\)\(-\frac{\sqrt{5}}{2}}{t_{2}+\frac{5}{2}+\frac{\sqrt{5}}{2}}\right|}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}, \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{14}{41}x-\frac{3}{41}\ln{|t_{1}|}+\frac{39}{41}.\frac{1}{\sqrt{5}}\ln{\left|\frac{2t_{2}+5-\sqrt{5}}{2t_{2}+5+\sqrt{5}}\right|}+c\) ➜Note লব ও হরের সহিত \(2\) গুণ করে।
\(=\frac{14}{41}x-\frac{3}{41}\ln{|t_{1}|}+\frac{39}{41\sqrt{5}}\ln{\left|\frac{2t_{2}+5-\sqrt{5}}{2t_{2}+5+\sqrt{5}}\right|}+c\)
\(=\frac{14}{41}x-\frac{3}{41}\ln{|4\cos{x}+5\sin{x}+6|}+\frac{39}{41\sqrt{5}}\ln{\left|\frac{2\tan{\left(\frac{x}{2}\right)}+5-\sqrt{5}}{2\tan{\left(\frac{x}{2}\right)}+5+\sqrt{5}}\right|}+c\) ➜Note \(\because t_{1}=4\cos{x}+5\sin{x}+6, t_{2}=\tan{\left(\frac{x}{2}\right)}\)
\(=\frac{14x}{41}-\frac{3}{41}\ln{|4\cos{x}+5\sin{x}+6|}+\frac{39}{41\sqrt{5}}\ln{\left|\frac{2\tan{\left(\frac{x}{2}\right)}\)\(+5-\sqrt{5}}{2\tan{\left(\frac{x}{2}\right)}+5+\sqrt{5}}\right|}+c\)

ধরি,
\(2\sin{x}+\cos{x}+3=L(\sin{x}+2\cos{x}+4)\)\(+M\left\{\frac{d}{dx}(\sin{x}+2\cos{x}+4)\right\}+N\)
\(\Rightarrow 2\sin{x}+\cos{x}+3=L(\sin{x}+2\cos{x}+4)+M(\cos{x}-2\sin{x})+N ....(1)\)
\(=L\sin{x}+2L\cos{x}+4L+M\cos{x}-2M\sin{x}+N\)
\(=(L-2M)\sin{x}+(2L+M)\cos{x}+4L+N\)
\(\therefore 2\sin{x}+\cos{x}+3=(L-2M)\sin{x}+(2L+M)\cos{x}+4L+N\)
\(\Rightarrow (L-2M)\sin{x}+(2L+M)\cos{x}+4L+N=2\sin{x}+\cos{x}+3 ........(2)\)
\(\Rightarrow L-2M=2 ....(3)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\sin{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 2L+M=1.....(4)\) ➜Note \((2)\) এর উভয় পার্শ হতে \(\cos{x}\) এর সহগ সমীকৃত করে ।
\(\Rightarrow 4L+N=3.....(5)\) ➜Note \((2)\) এর উভয় পার্শ হতে ধ্রুবক রাশির সমতা নিয়ে ।
\((3)\Rightarrow L-2M-2=0\)
\((4)\Rightarrow 2L+M-1=0\)
\(\Rightarrow \frac{L}{2+2}=\frac{M}{-4+1}=\frac{1}{1+4}\) ➜Note \((3)\) ও \((4)\) বজ্রগুণ করে।
\(\Rightarrow \frac{L}{4}=\frac{M}{-3}=\frac{1}{5}\)
\(\Rightarrow \frac{L}{4}=\frac{1}{5}; \frac{M}{-3}=\frac{1}{5}\)
\(\therefore L=\frac{4}{5}; M=-\frac{3}{5}\)
\(L\) এর মান \((5)\) এ বসিয়ে,
\(4.\frac{4}{5}+N=3\)
\(\Rightarrow \frac{16}{5}+N=3\)
\(\Rightarrow N=3-\frac{16}{5}\)
\(\Rightarrow N=\frac{15-16}{5}\)
\(\Rightarrow N=\frac{-1}{5}\)
\(\therefore N=-\frac{1}{5}\)
\(L\), \(M\) ও \(N\) এর মান \((1)\) এ বসিয়ে,
\(2\sin{x}+\cos{x}+3=\frac{4}{5}(\sin{x}+2\cos{x}+4)-\frac{3}{5}(\cos{x}-2\sin{x})-\frac{1}{5}\) \(\int{\frac{2\sin{x}+\cos{x}+3}{\sin{x}+2\cos{x}+4}dx}\)
\(=\int{\frac{\frac{4}{5}(\sin{x}+2\cos{x}+4)-\frac{3}{5}(\cos{x}-2\sin{x})-\frac{1}{5}}{\sin{x}+2\cos{x}+4}dx}\)
\(=\int{\left\{\frac{\frac{4}{5}(\sin{x}+2\cos{x}+4)}{\sin{x}+2\cos{x}+4}\)\(-\frac{\frac{3}{5}(\cos{x}-2\sin{x})}{\sin{x}+2\cos{x}+4}-\frac{\frac{1}{5}}{\sin{x}+2\cos{x}+4}\right\}dx}\)
\(=\frac{4}{5}\int{\frac{\sin{x}+2\cos{x}+4}{\sin{x}+2\cos{x}+4}dx}\)\(-\frac{3}{5}\int{\frac{\cos{x}-2\sin{x}}{\sin{x}+2\cos{x}+4}dx}-\frac{1}{5}\int{\frac{1}{\sin{x}+2\cos{x}+4}dx}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{\cos{x}-2\sin{x}}{\sin{x}+2\cos{x}+4}dx}\)\(-\frac{1}{5}\int{\frac{1}{\left(\frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)+2\left(\frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}\right)+4}dx}\) ➜Note \(\because \sin{A}=\frac{2\tan{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\), \(\cos{A}=\frac{1-\tan^2{\frac{A}{2}}}{1+\tan^2{\frac{A}{2}}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{\cos{x}-2\sin{x}}{\sin{x}+2\cos{x}+4}dx}\)\(-\frac{1}{5}\int{\frac{1+\tan^2{\frac{x}{2}}}{2\tan{\frac{x}{2}}+2-2\tan^2{\frac{x}{2}}+4+4\tan^2{\frac{x}{2}}}dx}\) ➜Note লব ও হরের সহিত \(1+\tan^2{\frac{x}{2}}\) গুণ করে।
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{\cos{x}-2\sin{x}}{\sin{x}+2\cos{x}+4}dx}\)\(-\frac{1}{5}\int{\frac{1+\tan^2{\frac{x}{2}}}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}+6}dx}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{\cos{x}-2\sin{x}}{\sin{x}+2\cos{x}+4}dx}\)\(-\frac{1}{5}\int{\frac{\sec^2{\left(\frac{x}{2}\right)}}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}+6}dx}\) ➜Note \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{\sin{x}+2\cos{x}+4}(\cos{x}-2\sin{x})dx}\)\(-\frac{1}{5}\int{\frac{1}{2\tan^2{\frac{x}{2}}+2\tan{\frac{x}{2}}+6}\sec^2{\left(\frac{x}{2}\right)}dx}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{2t^2_{2}+2t_{2}+6}2dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{2(t^2_{2}+t_{2}+3)}2dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{t^2_{2}+t_{2}+3}dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{t^2_{2}+2.t_{2}.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}+3}dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{\left(t_{2}+\frac{1}{2}\right)^2+3-\frac{1}{4}}dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{\left(t_{2}+\frac{1}{2}\right)^2+\frac{12-1}{4}}dt_{2}}\)
\(=\frac{4}{5}\int{dx}-\frac{3}{5}\int{\frac{1}{t_{1}}dt_{1}}-\frac{1}{5}\int{\frac{1}{\left(\frac{\sqrt{11}}{2}\right)^2+\left(t_{2}+\frac{1}{2}\right)^2}dt_{2}}\)
\(=\frac{4}{5}x-\frac{3}{5}\ln{|t_{1}|}-\frac{1}{5}.\frac{1}{\frac{\sqrt{11}}{2}}\tan^{-1}{\left(\frac{t_{2}+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}, \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{4}{5}x-\frac{3}{5}\ln{|t_{1}|}-\frac{1}{5}.\frac{2}{\sqrt{11}}\tan^{-1}{\left(\frac{2t_{2}+1}{\sqrt{11}}\right)}+c\) ➜Note লব ও হরের সহিত \(2\) গুণ করে।
\(=\frac{4x}{5}-\frac{3}{5}\ln{|t_{1}|}-\frac{2}{5\sqrt{11}}\tan^{-1}{\left[\frac{1}{\sqrt{11}}(2t_{2}+1)\right]}+c\)
\(=\frac{4x}{5}-\frac{3}{5}\ln{\left|\sin{x}+2\cos{x}+4\right|}\)\(-\frac{2}{5\sqrt{11}}\tan^{-1}{\left[\frac{1}{\sqrt{11}}\left\{2\tan{\left(\frac{x}{2}\right)}+1\right\}\right]}+c\) ➜Note \(\because t_{1}=\sin{x}+2\cos{x}+4, t_{2}=\tan{\left(\frac{x}{2}\right)}\)
\(=\frac{4x}{5}-\frac{3}{5}\ln{\left|\sin{x}+2\cos{x}+4\right|}\)\(-\frac{2}{5\sqrt{11}}\tan^{-1}{\left[\frac{1}{\sqrt{11}}\left\{1+2\tan{\left(\frac{x}{2}\right)}\right\}\right]}+c\)

\(\int{\frac{dx}{(x-1)\sqrt{x+4}}}\)
\(=\int{\frac{1}{(x-1)\sqrt{x+4}}dx}\)
\(=\int{\frac{1}{(t^2-4-1)t}2tdt}\) ➜Note \(\because x=t^2-4\)
\(=2\int{\frac{1}{t^2-4-1}dt}\)
\(=2\int{\frac{1}{t^2-5}dt}\)
\(=2\int{\frac{1}{t^2-(\sqrt{5})^2}dt}\)
\(=2.\frac{1}{2.\sqrt{5}}\ln{\left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{5}}\ln{\left|\frac{\sqrt{x+4}-\sqrt{5}}{\sqrt{x+4}+\sqrt{5}}\right|}+c\) ➜Note \(\because t=\sqrt{x+4}\)

\(\int{\frac{dx}{(x-3)\sqrt{x+1}}}\)
\(=\int{\frac{1}{(x-3)\sqrt{x+1}}dx}\)
\(=\int{\frac{1}{(t^2-1-3)t}2tdt}\) ➜Note \(\because x=t^2-1\)
\(=2\int{\frac{1}{t^2-1-3}dt}\)
\(=2\int{\frac{1}{t^2-4}dt}\)
\(=2\int{\frac{1}{t^2-2^2}dt}\)
\(=2.\frac{1}{2.2}\ln{\left|\frac{t-2}{t+2}\right|}+c\) ➜Note \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sqrt{2}}\ln{\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|}+c\) ➜Note \(\because t=\sqrt{x+1}\)

\(\int{\frac{dx}{(1-x)\sqrt{1-x^2}}}\)
\(=\int{\frac{1}{(1-x)\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\frac{1}{t}\sqrt{1-\left(1-\frac{1}{t}\right)^2}}.\frac{1}{t^2}dt}\) ➜Note \(\because x=1-\frac{1}{t}\)
\(=\int{\frac{1}{\sqrt{1-\left(1-2.1.\frac{1}{t}+\frac{1}{t^2}\right)}}.\frac{1}{t}dt}\)
\(=\int{\frac{1}{\sqrt{1-1+\frac{2}{t}-\frac{1}{t^2}}}.\frac{1}{t}dt}\)
\(=\int{\frac{1}{\sqrt{\frac{2}{t}-\frac{1}{t^2}}}.\frac{1}{t}dt}\)
\(=\int{\frac{1}{\sqrt{\frac{2t-1}{t^2}}}.\frac{1}{t}dt}\)
\(=\int{\frac{1}{\frac{1}{t}.\sqrt{2t-1}}.\frac{1}{t}dt}\)
\(=\int{\frac{1}{\sqrt{2t-1}}dt}\)
\(=\int{\frac{1}{z}zdz}\)
\(=\int{dz}\)
\(=z+c\) ➜Note \(\because \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sqrt{2t-1}+c\) ➜Note \(\because z=\sqrt{2t-1}\)
\(=\sqrt{2\left(\frac{1}{1-x}\right)-1}+c\) ➜Note \(\because 1-x=\frac{1}{t}\Rightarrow t=\frac{1}{1-x}\)
\(=\sqrt{\frac{2}{1-x}-1}+c\)
\(=\sqrt{\frac{2-1+x}{1-x}}+c\)
\(=\sqrt{\frac{1+x}{1-x}}+c\)

\(\int{\sqrt{1+\sec{x}}dx}\)
\(=\int{\sqrt{1+\frac{1}{\cos{x}}}dx}\)
\(=\int{\sqrt{\frac{\cos{x}+1}{\cos{x}}}dx}\)
\(=\int{\frac{\sqrt{1+\cos{x}}}{\sqrt{\cos{x}}}dx}\)
\(=\int{\frac{\sqrt{2\cos^2{\left(\frac{x}{2}\right)}}}{\sqrt{1-2\sin^2{\left(\frac{x}{2}\right)}}}dx}\)
\(=\sqrt{2}\int{\frac{\cos{\left(\frac{x}{2}\right)}}{\sqrt{1-2\sin^2{\left(\frac{x}{2}\right)}}}dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{2}\sqrt{\frac{1}{2}-\sin^2{\left(\frac{x}{2}\right)}}}\cos{\left(\frac{x}{2}\right)}dx}\)
\(=\int{\frac{1}{\sqrt{\frac{1}{2}-t^2}}.2dt}\)
\(=2\int{\frac{1}{\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2-t^2}}dt}\)
\(=2\sin^{-1}{\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sin^{-1}{\left(\sqrt{2}t\right)}+c\)
\(=2\sin^{-1}{\left\{\sqrt{2}\sin{\left(\frac{x}{2}\right)}\right\}}+c\) ➜Note \(\because t=\sin{\left(\frac{x}{2}\right)}\)

\(\int{\cos{\left\{2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right\}}dx}\)
\(=\int{\cos{\left\{2\cot^{-1}\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\right\}}\times{-\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left\{2\cot^{-1}\sqrt{\frac{2\sin^2{\left(\frac{\theta}{2}\right)}}{2\cos^2{\left(\frac{\theta}{2}\right)}}}\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left\{2\cot^{-1}\sqrt{\frac{\sin^2{\left(\frac{\theta}{2}\right)}}{\cos^2{\left(\frac{\theta}{2}\right)}}}\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left[2\cot^{-1}\left\{\sqrt{\tan^2{\left(\frac{\theta}{2}\right)}}\right\}\right]}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left[2\cot^{-1}\left\{\tan{\left(\frac{\theta}{2}\right)}\right\}\right]}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left[2\cot^{-1}\left\{\cot{\left(\frac{\pi}{2}-\frac{\theta}{2}\right)}\right\}\right]}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left\{2\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\cos{\left(\pi-\theta\right)}\times{\sin{\theta}d\theta}}\)
\(=-\int{(-\cos{\theta})\sin{\theta}d\theta}\)
\(=\int{\cos{\theta}\sin{\theta}d\theta}\)
\(=\frac{1}{2}\int{2\sin{\theta}\cos{\theta}d\theta}\)
\(=\frac{1}{2}\int{\sin{2\theta}d\theta}\)
\(=\frac{1}{2}\int{\sin{t}.\frac{1}{2}dt}\)
\(=\frac{1}{4}\int{\sin{t}dt}\)
\(=\frac{1}{4}(-\cos{t})+c_{1}\) ➜Note \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\cos{t}+c_{1}\)
\(=-\frac{1}{4}\cos{2\theta}+c_{1}\) ➜Note \(\because t=2\theta\)
\(=-\frac{1}{4}\left(2\cos^2{\theta}-1\right)+c_{1}\) ➜Note \(\because \cos{2A}=2\cos^2{A}-1\)
\(=-\frac{1}{4}\left(2x^2-1\right)+c_{1}\) ➜Note \(\because \cos{\theta}=x\)
\(=-\frac{1}{2}x^2+\frac{1}{4}+c_{1}\)
\(=-\frac{1}{2}x^2+c\) ➜Note \(\because \frac{1}{4}+c_{1}=c\)

\(\int{\sin{\left\{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\}}dx}\)
\(=\int{\sin{\left\{2\tan^{-1}\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\right\}}\times{-\sin{\theta}d\theta}}\)
\(=-\int{\sin{\left\{2\tan^{-1}\sqrt{\frac{2\sin^2{\left(\frac{\theta}{2}\right)}}{2\cos^2{\left(\frac{\theta}{2}\right)}}}\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin{\left\{2\tan^{-1}\sqrt{\frac{\sin^2{\left(\frac{\theta}{2}\right)}}{\cos^2{\left(\frac{\theta}{2}\right)}}}\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin{\left[2\tan^{-1}\left\{\sqrt{\tan^2{\left(\frac{\theta}{2}\right)}}\right\}\right]}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin{\left[2\tan^{-1}\left\{\tan{\left(\frac{\theta}{2}\right)}\right\}\right]}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin{\left\{2\left(\frac{\theta}{2}\right)\right\}}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin{\theta}\times{\sin{\theta}d\theta}}\)
\(=-\int{\sin^2{\theta}d\theta}\)
\(=-\frac{1}{2}\int{2\sin^2{\theta}d\theta}\)
\(=-\frac{1}{2}\int{(1-\cos{2\theta})d\theta}\) ➜Note \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=-\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{2\theta}d\theta}\)
\(=-\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{t}.\frac{1}{2}dt}\)
\(=-\frac{1}{2}\int{d\theta}+\frac{1}{4}\int{\cos{t}dt}\)
\(=-\frac{1}{2}\theta+\frac{1}{4}\sin{t}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\theta+\frac{1}{4}\sin{(2\theta)}+c\) ➜Note \(\because t=2\theta\)
\(=-\frac{1}{2}\theta+\frac{1}{4}.2\sin{\theta}\cos{\theta}+c\) ➜Note \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=-\frac{1}{2}\theta+\frac{1}{2}\sin{\theta}\cos{\theta}+c\)
\(=-\frac{1}{2}\theta+\frac{1}{2}\sqrt{1-\cos^2{\theta}}\cos{\theta}+c\)
\(=-\frac{1}{2}\theta+\frac{1}{2}\cos{\theta}\sqrt{1-\cos^2{\theta}}+c\)
\(=-\frac{1}{2}\cos^{-1}{x}+\frac{1}{2}x\sqrt{1-x^2}+c\) ➜Note \(\because \cos{\theta}=x, \theta=\cos^{-1}{x}\)

\(\int{\frac{dx}{x+\sqrt{x}}}\)
\(=\int{\frac{1}{(\sqrt{x})^2+\sqrt{x}}dx}\)
\(=\int{\frac{1}{t^2+t}.2tdt}\)
\(=2\int{\frac{t}{t(t+1)}dt}\)
\(=2\int{\frac{1}{t+1}dt}\)
\(=2\int{\frac{1}{z}dz}\)
\(=2\ln{|z|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{|t+1|}+c\) ➜Note \(\because z=t+1\)
\(=2\ln{(\sqrt{x}+1)}+c\) ➜Note \(\because t=\sqrt{x}\)

\(\int{x\sqrt{\frac{1-x}{1+x}}dx}\)
\(=\int{x\frac{\sqrt{1-x}}{\sqrt{1+x}}dx}\)
\(=\int{x\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1+x}.\sqrt{1-x}}dx}\) ➜Note লব ও হরের সহিত \(\sqrt{1-x}\) গুণ করে।
\(=\int{\frac{x(1-x)}{\sqrt{(1+x)(1-x)}}dx}\)
\(=\int{\frac{x-x^2}{\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1-x^2+x-1}{\sqrt{1-x^2}}dx}\)
\(=\int{\left(\frac{1-x^2}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\left(\frac{\sqrt{1-x^2}.\sqrt{1-x^2}}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\left(\sqrt{1-x^2}+\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\right)dx}\)
\(=\int{\sqrt{1-x^2}dx}+\int{\frac{1}{\sqrt{1-x^2}}xdx}-\int{\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{\sqrt{1-x^2}dx}+\int{\frac{1}{t}\times{-tdt}}-\int{\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{\sqrt{1-x^2}dx}-\int{dt}-\int{\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}\sin^{-1}{x}-t-\sin^{-1}{x}+c\) ➜Note \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\), \(\int{dx}=x, \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}\sin^{-1}{x}-\sqrt{1-x^2}-\sin^{-1}{x}+c\) ➜Note \(\because t=\sqrt{1-x^2}\)
\(=\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}\sin^{-1}{x}-\sin^{-1}{x}-\sqrt{1-x^2}+c\)
\(=\frac{x\sqrt{1-x^2}}{2}+\frac{1-2}{2}\sin^{-1}{x}-\sqrt{1-x^2}+c\)
\(=\frac{x\sqrt{1-x^2}}{2}+\frac{-1}{2}\sin^{-1}{x}-\sqrt{1-x^2}+c\)
\(=\frac{x\sqrt{1-x^2}}{2}-\frac{1}{2}\sin^{-1}{x}-\sqrt{1-x^2}+c\)

\(\int{\sin^{-1}{\sqrt{\frac{x}{a+x}}}dx}\)
\(=\int{\sin^{-1}{\sqrt{\frac{a\tan^2{\theta}}{a+a\tan^2{\theta}}}}\times{2a\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\sqrt{\frac{a\tan^2{\theta}}{a(1+\tan^2{\theta})}}}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\sqrt{\frac{\tan^2{\theta}}{1+\tan^2{\theta}}}}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\sqrt{\frac{\tan^2{\theta}}{\sec^2{\theta}}}}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\left(\frac{\tan{\theta}}{\sec{\theta}}\right)}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\left(\frac{\frac{\sin{\theta}}{\cos{\theta}}}{\frac{1}{\cos{\theta}}}\right)}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{\left(\frac{\sin{\theta}}{\cos{\theta}}\times{\cos{\theta}}\right)}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\sin^{-1}{(\sin{\theta})}\times{\tan{\theta}\sec^2{\theta}d\theta}}\)
\(=2a\int{\theta\tan{\theta}\sec^2{\theta}d\theta}\)
\(=2a\int{\tan^{-1}{t}.t.dt}\)
\(=2a\int{t\tan^{-1}{t}dt}\)
\(=2a\left[\tan^{-1}{t}\int{tdt}-\int{\left\{\frac{d}{dt}(\tan^{-1}{t})\int{tdt}\right\}dt}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=2a\left[\frac{t^2}{2}\tan^{-1}{t}-\int{\frac{1}{1+t^2}.\frac{t^2}{2}dt}\right]\)
\(=2a\left[\frac{t^2}{2}\tan^{-1}{t}-\frac{1}{2}\int{\frac{t^2}{1+t^2}dt}\right]\)
\(=at^2\tan^{-1}{t}-a\int{\frac{1+t^2-1}{1+t^2}dt}\)
\(=at^2\tan^{-1}{t}-a\int{\left(\frac{1+t^2}{1+t^2}-\frac{1}{1+t^2}\right)dt}\)
\(=at^2\tan^{-1}{t}-a\int{\left(1-\frac{1}{1+t^2}\right)dt}\)
\(=at^2\tan^{-1}{t}-a\int{dt}+a\int{\frac{1}{1+t^2}dt}\)
\(=at^2\tan^{-1}{t}-at+a\tan^{-1}{t}+c\) ➜Note \(\because \int{dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=a\tan^2{\theta}.\theta-a\tan{\theta}+a\theta+c\) ➜Note \(\because t^2=\tan^2{\theta}, \tan^{-1}{t}=\theta, t=\tan{\theta}\)
\(=a.\frac{x}{a}.\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-a\sqrt{\frac{x}{a}}+a\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}+c\) ➜Note \(\because \tan^2{\theta}=\frac{x}{a}, \theta=\tan^{-1}{\sqrt{\frac{x}{a}}}, \tan{\theta}=\sqrt{\frac{x}{a}}\)
\(=x\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{a}.\sqrt{a}\frac{\sqrt{x}}{\sqrt{a}}+a\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}+c\)
\(=a\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}+x\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{a}\sqrt{x}+c\)
\(=a\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}+x\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{(ax)}+c\)
\(=(a+x)\tan^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{(ax)}+c\)