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? 
 A: Hint:
Let $f(u) = u$ and  $g(u) = 1-u^2$. Hence,
$$ \left(\frac{f}{g}\right)'(u) = \frac{u^2+1}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} +  \frac{1-u^2}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} + \frac{1}{2} \left(\frac{1}{1-u} + \frac{1}{1+u}\right).$$
A: Hint:
$$\frac{u^2}{(1-u^2)^2} = \frac{u^2-1+1}{(1-u^2)^2} = \frac{u^2-1}{(1-u^2)^2}+\frac{1}{(1-u^2)^2}=\frac{1}{(1-u^2)^2}-\frac{1}{1-u^2}.$$
then see your previous post and you'll get the result, 
A: Hint: $u = \sin \theta$. 
$$...=\int \frac{\sin^2\theta\cos\theta}{\cos^4\theta}d\theta = \int \tan^2\theta \sec\theta d\theta$$
A: 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,

A: The $u=\tanh t$ substitution is probably the shortest way: $\,\, \mathrm d\,u=\dfrac{\mathrm d\,t}{\cosh t}$, so that
$$ \int\frac{u^2}{(1-u^2)^2} \mathrm d\,u=\int \sinh^2t\,\mathrm d\,t=\int\dfrac{\cosh2t-1}2 \mathrm d\,t=\frac14\sinh 2t - \frac t2. $$
Now $\,\,\sinh 2t =\dfrac{2\tanh t}{1-\tanh^2t}=\dfrac{2u}{1-u^2}\,$, so
$$ \int\frac{u^2}{(1-u^2)^2} \mathrm d\,u=\frac 12\dfrac{u}{1-u^2}-\frac14\ln\biggl(\frac{1+u}{1-u}\biggr).$$
