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

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