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? 

For the diversity of the solution set, I present an exotic method that works almost exclusively for this integral:
Decompose the integrand as
\begin{align*}
\frac{u^{2}}{(u^{2}-1)^{2}}
&= \frac{1}{u^{2} + u^{-2} - 2} \\
&= \frac{1}{2}\frac{1+u^{-2}}{(u-u^{-1})^{2}} + \frac{1}{2}\frac{1-u^{-2}}{(u+u^{-1})^{2} - 4} \\
&= \frac{1}{2}\frac{1+u^{-2}}{(u-u^{-1})^{2}} + \frac{1}{8} \frac{1-u^{-2}}{(u+u^{-1}) - 2} - \frac{1}{8} \frac{1-u^{-2}}{(u+u^{-1}) + 2}.
\end{align*}
By noting that
$$ \frac{d}{du}(u-u^{-1}) = 1+u^{-2}
\qquad \text{and} \qquad
\frac{d}{du}(u+u^{-1}) = 1-u^{-2}, $$
it follows that
\begin{align*}
\int \frac{u^{2}}{(u^{2}-1)^{2}} \, du
&= - \frac{1}{2}\frac{1}{u-u^{-1}} + \frac{1}{8} \log \left|(u+u^{-1}) - 2\right| - \frac{1}{8} \log \left|(u+u^{-1}) + 2\right| \\
&= \frac{u}{2(1-u^{2})} + \frac{1}{4} \log \left| \frac{1-u}{1+u} \right|.
\end{align*}
Indeed,

