\(L.S=\int{\frac{1}{a^2+x^2}dx}\)
\(=\int{\frac{1}{a^2+a^2\tan^2{\theta}}\times{a\sec^2{\theta}}d\theta}\)
\(=\int{\frac{1}{a^2(1+\tan^2{\theta})}\times{a\sec^2{\theta}}d\theta}\)
\(=\int{\frac{1}{a\sec^2{\theta}}\times{\sec^2{\theta}}d\theta}\) ➜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 \(\because x-b=t(x-a)\Rightarrow \frac{x-b}{x-a}=t\)
\(=\frac{1}{(a-b)^4}\left[-\frac{(x-a)^2}{2(x-b)^2}+\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{x-b}{x-a}\right]+c\)
\(=\frac{1}{(a-b)^4}\left[\frac{3(x-a)}{x-b}+3\ln{\left|\frac{x-b}{x-a}\right|}-\frac{(x-a)^2}{2(x-b)^2}-\frac{x-b}{x-a}\right]+c\)