Question

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

Answer 1

$\int \frac{2}{\left(1-x\right)\left(1+x^2\right)}dx$

$\frac{2}{\left(1-x\right)\left(1+x^2\right)}=\frac{A}{1-x}+\frac{Bx+c}{x^2+1}$

$=\frac{2}{\left(1-x\right)\left(1+x^2\right)}=\frac{A\left(x^2+1\right)+\left(B+c\right)\left(1-x\right)}{\left(1-x\right)\left(x^2+1\right)}$

On Comparing both side we get

$A-B=0$ $B-C=0$ $A+C=2$

$\Rightarrow A=B$ $\Rightarrow B=C$

$A+C=2$

$\Rightarrow A+A=2$

$\Rightarrow 2A=2$

$\Rightarrow A=1$

So $A=B=C=1$

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

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

$=-\log \left|x-1\right|+\frac{1}{2}\log \left(x^2+1\right)+{\tan }^{-1}x+c$

$=\frac{1}{2}\left[\log \left(x^2+1\right)-2\log \left|x-1\right|+2{\tan }^{-1}x\right]+c$

$=\frac{1}{2}\left[\log \frac{x^2+1}{{\left(x-1\right)}^2}+2{\tan }^{-1}x\right]+c$

$=\frac{1}{2}\left[\log \frac{x^2+1}{{\left(1-x\right)}^2}+2{\tan }^{-1}x\right]+c$