TITLE: Hints to integrate $\int \frac{1+t^2}{(1-t^2)^2} dt$
QUESTION [8 upvotes]: Integrate $\int \frac{1+t^2}{(1-t^2)^2} dt \\$
I just want some hint, not the full answer. I've tried substituting $v= 1\pm t^2$ and I've tried  partial fraction but I end up with weird results. Any help?

REPLY [2 votes]: Maybe use
$\dfrac{1+t^2}{(1-t^2)^2} = \dfrac{1}{2}\dfrac { (1-t)^2 + (1+t)^2}{((1-t)(1+t))^2}$
so you end up integrating $\dfrac{1}{2}\cdot \left( \dfrac{1}{(1+t)^2} + \dfrac{1}{(1-t)^2} \right)$
where each is easy with substitution $1+t = u, 1-t = v$