Find: ∫ 2/((1-x)(1+x^2)) d x

$$\text { Let } \frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}} $$ $$\Rightarrow A=1, B=1, C=1 $$ Hence, $\displaystyle \mathrm{I}=\int \frac{2}{(1-x)\left(1-x^{2}\right)} \mathrm{dx}=\int\left[\frac{1}{1-x}+\frac{x+1}{1+x^{2}}\right] \mathrm{dx}$ $$\begin{aligned} & =\int \frac{1}{1-x} d x+\int \frac{x+1}{x^{2}+1} d x \\ & =\int \frac{1}{1-x} d x+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x+\int \frac{1}{x^{2}+1} d x \end{aligned} $$ $$=-\log |1-x|+\frac{1}{2} \log \left(x^{2}+1\right)+\tan ^{-1}(x)+C $$