Question:
$\frac{2}{(1-x)\left(1+x^{2}\right)}$
Solution:
Let $\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{(1-x)}+\frac{B x+C}{\left(1+x^{2}\right)}$
$2=A\left(1+x^{2}\right)+(B x+C)(1-x)$
$2=A+A x^{2}+B x-B x^{2}+C-C x$
Equating the coefficient of $x^{2}, x$, and constant term, we obtain
$A-B=0$
$B-C=0$
$A+C=2$
On solving these equations, we obtain
$A=1, B=1$, and $C=1$
$\therefore \frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^{2}}$
$\Rightarrow \int \frac{2}{(1-x)\left(1+x^{2}\right)} d x=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$
$=-\int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$
$=-\log |x-1|+\frac{1}{2} \log \left|1+x^{2}\right|+\tan ^{-1} x+\mathrm{C}$