মূলদ ভগ্নাংশের যোগজীকরণ ( Integration of Rational Fractions )

অনুশীলনী \(10.E / Q.2\)-এর প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(Q.2.(i)\) \(\int{\frac{1}{x(x+1)^2}dx}\)
উত্তরঃ \(\ln{\left|\frac{x}{x+1}\right|}+\frac{1}{x+1}+c\)

\(Q.2.(ii)\) \(\int{\frac{x^2dx}{(x+1)(x+2)^2}}\)
উত্তরঃ \(\ln{|x+1|}+\frac{4}{x+2}+c\)

\(Q.2.(iii)\) \(\int{\frac{5-2x}{(x-1)^2(x+2)}dx}\)
উত্তরঃ \(\ln{\left|\frac{x+2}{x-1}\right|}-\frac{1}{x-1}+c\)

\(Q.2.(iv)\) \(\int{\frac{1}{x^2(x-1)}dx}\)
উত্তরঃ \(\ln{\left|\frac{x-1}{x}\right|}+\frac{1}{x}+c\)
[ বঃ২০১০,২০০৫; রাঃ২০০২; কুঃ২০০২ ]

\(Q.2.(v)\) \(\int{\frac{x}{(x-1)^2(x+2)}dx}\)
উত্তরঃ \(\frac{2}{9}\ln{\left|\frac{x-1}{x+2}\right|}-\frac{1}{3(x-1)}+c\)

\(Q.2.(vi)\) \(\int{\frac{xdx}{(x-1)(x^2+1)}}\)
উত্তরঃ \(\frac{1}{2}\ln{|x-1|}-\frac{1}{4}\ln{|x^2+1|}+\frac{1}{2}\tan^{-1}{x}+c\)
[ ঢাঃ২০১৭,২০০৮; কুঃ২০১১,২০০৮; দিঃ২০০৯; বঃ২০০৭,২০০১ ]

\(Q.2.(vii)\) \(\int{\frac{dx}{x(x^2+1)}}\)
উত্তরঃ \(\ln{|x|}-\frac{1}{2}\ln{|x^2+1|}+c\)
\(Q.2.(viii)\) \(\int{\frac{x^2dx}{x^4-1}}\)
উত্তরঃ \(\frac{1}{4}\ln{\left|\frac{x-1}{x+1}\right|}+\frac{1}{2}\tan^{-1}{x}+c\)

\(Q.2.(ix)\) \(\int{\frac{2x+3}{x^3-x}dx}\)
উত্তরঃ \(\frac{1}{2}\ln{|x+1|}-3\ln{|x|}+\frac{5}{2}\ln{|x-1|}+c\)

\(Q.2.(x)\) \(\int{\frac{x+2}{(1-x)(x^2+4)}dx}\)
উত্তরঃ \(\frac{3}{5}\ln{|1-x|}+\frac{3}{10}\ln{|x^2+4|}-\frac{1}{5}\tan^{-1}{\left(\frac{x}{2}\right)}+c\)

\(Q.2.(xi)\) \(\int{\frac{x}{(x-1)(x^2+4)}dx}\)
উত্তরঃ \(\frac{1}{5}\ln{|x-1|}-\frac{1}{10}\ln{|x^2+4|}+\frac{2}{5}\tan^{-1}{\left(\frac{x}{2}\right)}+c\)

\(Q.2.(xii)\) \(\int{\frac{dx}{x^2(x+1)^2}}\)
উত্তরঃ \(2\ln{\left|\frac{x+1}{x}\right|}-\frac{1}{x}-\frac{1}{x+1}+c\)

\(Q.2.(xiii)\) \(\int{\frac{2x^2}{(x^2+1)(x^2+3)}dx}\)
উত্তরঃ \(\sqrt{3}\tan^{-1}{\left(\frac{x}{\sqrt{3}}\right)}-\tan^{-1}{x}+c\)
1 2 3 4 5

Leave a Reply