If int frac{d x}{(x+2)left(x^2+1right)}=a log | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,$\int \frac{d x}{(x+2)\left(x^2+1ight)}=a \log \left|1+x^2ight|+b 	an ^{-1} x+\frac{1}{5} \log |x+2|+c$. Let $\frac{1}{(x+2)\left(x^2+1

Vedclass · Accepted Answer

$a=\frac{-1}{10}, b=\frac{2}{5}$

Vedclass · Answer

(A) Given,$\int \frac{d x}{(x+2)\left(x^2+1ight)}=a \log \left|1+x^2ight|+b 	an ^{-1} x+\frac{1}{5} \log |x+2|+c$. Let $\frac{1}{(x+2)\left(x^2+1ight)}=\frac{A}{(x+2)}+\frac{B x+C}{\left(x^2+1ight)}$. Then $1=A(x^2+1)+(x+2)(Bx+C) = (A+B)x^2 + (2B+C)x + (A+2C)$. Comparing coefficients,we get $A+B=0$,$2B+C=0$,and $A+2C=1$. Solving these,we find $A=\frac{1}{5}$,$B=-\frac{1}{5}$,and $C=\frac{2}{5}$. Thus,$\int \frac{d x}{(x+2)\left(x^2+1ight)} = \int \left( \frac{1}{5(x+2)} - \frac{x}{5(x^2+1)} + \frac{2}{5(x^2+1)} ight) dx$. $= \frac{1}{5} \log |x+2| - \frac{1}{10} \log |x^2+1| + \frac{2}{5} 	an^{-1} x + c$. Comparing this with the given expression,we get $a=-\frac{1}{10}$ and $b=\frac{2}{5}$.

If $\int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$,then

Similar Questions

$\int \frac{e^x}{(2+e^x)(e^x+1)} dx =$ (where $C$ is a constant of integration.)

Find $\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$

$\int \frac{x \, dx}{(x-1)^2(x+2)} = $

$\int \frac{1}{x - x^3} \, dx = $

If $\int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx = {{\log }_e}\left\{ {{{(x - 1)}^{\frac{5}{2}}}{{({x^2} + 1)}^a}} \right\}} - \frac{1}{2}{\tan ^{ - 1}}x + A$,where $A$ is any arbitrary constant,then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module