Question

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

Answer 1

$\int \frac{2}{(1-x)(1+x^2)}=$ $\int \frac{A}{(1-x)}$$+\int \frac{Bx+C}{(1+x^2)}$

$2=A(1+x^2)+(Bx+C)(1-x)$ .........(1)

put $x=1$ in eq.1

$2=A(1+1^2)+\left(B\right(1)+C)(1-1)$

$2A=2$

$A=1$

and now put $x=0andA=1$ in eq. (1)

$2=1(1+0)+\left(B\right(0)+C)(1-0)$

$C=1$

put value of $C$ and $A$ and $x=-1$ in eq. (1)

$2=\left(1\right)(1+(-1{)}^2)+(B(-1)+1\left)\right(1-(-1))$

$2=2+2(1-B)$

$B=1$

now,

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

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

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

$=$ $-log(1-x)+\frac{1}{2}log(1+x^2)+$ ${\tan }^{-1}x+C$