integration of $\large\int \frac{u^2}{(1-u^2)^2}$ $ du$ Is there a way to integrate $$\large\int \frac{u^2}{(1-u^2)^2} du$$ without using partial fraction decomposition? 

Hint:
Let $f(u) = u$ and  $g(u) = 1-u^2$. Hence,
$$ \left(\frac{f}{g}\right)'(u) = \frac{u^2+1}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} +  \frac{1-u^2}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} + \frac{1}{2} \left(\frac{1}{1-u} + \frac{1}{1+u}\right).$$
