How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$?

Question

When integrating $\int \frac{1-x}{x^2-1} dx$ Maple rewrote it as $-\int\frac{1}{x+1}dx$

How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$?

Answer 1

hint:

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

Answer 2

$\dfrac{1-x}{x^2-1} = \dfrac{-(x-1)}{(x+1)(x-1)}$ and if $ x \ne 1 $, then $\dfrac{-1}{x+1}$.

With this, $\int\dfrac{1-x}{x^2-1}dx = -\int\dfrac{1}{x+1}=- \ln|x+1|+C$.