Question

Evaluate

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

Answer 1

$\begin{array}{c}{\int }^{}\frac{1}{1-x^2}dx\\ =-{\int }^{}\frac{1}{x^2-1}dx\\ =-{\int }^{}\frac{1}{\left(x-1\right)\left(x+1\right)}dx\\ =-{\int }^{}\left(\frac{1}{2\left(x-1\right)}-\frac{1}{2\left(x+1\right)}\right)dx\\ =\frac{\ln \left(x+1\right)}{2}-\frac{\ln \left(x-1\right)}{2}+C\end{array}$