Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$ 
Calculate  $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$

I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$
$$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$
How can I solve it?
 A: You're definetly on the right track with that substitution of $x=\frac1t$ 
Basically we have: $$I=\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx=\int_0^\infty \frac{x^2\ln^2 x}{(1-x^2)^2}dx$$
Now what if we add them up?
$$2I=\int_0^\infty \ln^2 x \frac{1+x^2}{(1-x^2)^2}dx$$
If you don't know how to deal easily with the integral  $$\int \frac{1+x^2}{(1-x^2)^2}dx=\frac{x}{1-x^2}+C$$
I recommend you to take a look here.
Anyway we have, integrating by parts: 
$$2I= \underbrace{\frac{x}{1-x^2}\ln^2x \bigg|_0^\infty}_{=0} +2\underbrace{\int_0^\infty \frac{\ln x}{x^2-1}dx}_{\large =\frac{\pi^2}{4}}$$
$$\Rightarrow 2I= 2\cdot \frac{\pi^2}{4} \Rightarrow I=\frac{\pi^2}{4}$$
For the last integral see here for example.
A: This is a variant of TheSimpliFire's approach given in a comment. 
By letting $x=e^t$ we get 
$$\begin{align*}
\int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}\,dx
&=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\
&=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,dt\\
&=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+1)t}\,dt+\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\
&=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+1)^3}+\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}\\
&=\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)+\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}-\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)\\
&=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=2\left(\sum_{n=0}^\infty\frac{1}{n^2}-\sum_{n=0}^\infty\frac{1}{(2n)^2}\right)\\&=2\left(1-\frac{1}{4}\right)\sum_{n=0}^\infty\frac{1}{n^2}=\frac{3}{2}\cdot \frac{\pi^2}{6}=\frac{\pi^2}{4}.
\end{align*}$$
A: Note
\begin{align}
\int^{\infty}_{0}\frac{\ln^2x}{(1-x^2)^2} dx
&=\int^{\infty}_{0}\frac{\ln^2x}{2x} d\left(\frac{x^2}{1-x^2}\right)\\
&\overset{IBP}= -\int^{\infty}_{0}\frac{\ln x}{1-x^2} dx +\frac12 \int^{\infty}_{0}\frac{\ln^2x}{1-x^2} dx \\
&= -(-\frac{\pi^2}4)+\frac12\cdot 0=\frac{\pi^2}4
\end{align}
$\int^{\infty}_{0}\frac{\ln x}{1-x^2} dx =-\frac{\pi^2}4$
A: Here is yet another slight variation on a theme.
Let
$$I = \int_0^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx$$
then
\begin{align}
I &= \int_0^1 \frac{\ln^2 x}{(1 - x^2)^2} \, dx + \int_1^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx = \int_0^1 \frac{(1 + x^2) \ln^2 x}{(1 - x^2)^2} \, dx \tag1,
\end{align}
after a substitution of $x \mapsto 1/x$ has been enforced in the second of the integrals.
As
$$\frac{1}{1 - x^2} = \sum_{n = 0}^\infty x^{2n}, \qquad |x| < 1,$$
differentiating with respect to $x$ gives
$$\frac{1}{(1 - x^2)^2} = \sum_{n = 1}^\infty n x^{2n - 2}.$$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$$I = \sum_{n = 1}^\infty n \int_0^1 (x^{2n - 2} + x^{2n}) \ln^2 x \, dx.$$
Integrating by parts twice, we are left with
$$I = \sum_{n = 1}^\infty \left (\frac{2n}{(2n - 1)^3} + \frac{2n}{(2n + 1)^3} \right ).$$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
\begin{align}
I &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} + \frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= \sum_{n = 0}^\infty \left [\frac{1}{(2n + 1)^2} + \frac{1}{(2n + 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\
&= 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n + 1)^2}\\
&= 2 \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}\\
&= 2 \left [\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} \right ]\\
&= 2 \left (1 - \frac{1}{4} \right ) \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= \frac{3}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= \frac{3}{2} \cdot \frac{\pi^2}{6}\\
&= \frac{\pi^2}{4},
\end{align}
as expected.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x}
\over \pars{1 - x^{2}}^{2}}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 8}\int_{0}^{\infty}{x^{-1/2}\,\ln^{2}\pars{x}
\over \pars{1 - x}^{2}}\,\dd x
\\[5mm] = &\
\left.{1 \over 8}\partiald[2]{}{\nu}\int_{0}^{\infty}{x^{-1/2}\pars{x^{\nu} - 1}
\over \pars{1 - x}^{2}}\,\dd x
\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] = &\
\left.{1 \over 8}\partiald[2]{}{\nu}
\int_{0}^{\infty}
{x^{\pars{\nu + 1/2} - 1}\,\,\, - x^{1/2 - 1} \over \pars{1 - x}^{2}}\,\dd x
\,\right\vert_{\ \nu\ =\ 0}
\end{align}

\begin{align}
&{1 \over \pars{1 - x}^{2}} =
\sum_{k = 0}^{\infty}{-2 \choose k}\pars{-x}^{k} =
\sum_{k = 0}^{\infty}{k + 1 \choose k}\pars{-1}^{k}
\pars{-x}^{k}
\\[5mm] = &\
\sum_{k = 0}^{\infty}
\color{red}{\pars{1 + k}\Gamma\pars{1 + k}\expo{\ic\pi k}}{\pars{-x}^{k} \over k!}
\end{align}

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x}
\over \pars{1 - x^{2}}^{2}}\,\dd x} =
{1 \over 8}\partiald[2]{}{\nu}
\bracks{\Gamma\pars{\nu + {1 \over 2}}
\pars{{1 \over 2} - \nu}\Gamma\pars{{1 \over 2} - \nu}
\expo{-\ic\pi\pars{\nu + 1/2}}}_{\ \nu\ =\ 0}
\\[5mm] = &\
-\,{\pi \over 8}\ic\,\partiald[2]{}{\nu}
\bracks{\pars{{1 \over 2} - \nu}\bracks{1 -
\ic\tan\pars{\pi\nu}}}_{\ \nu\ =\ 0}
\\[5mm] = &\
{\pi \over 8}\,\partiald[2]{}{\nu}
\bracks{\pars{\nu - {1 \over 2}}\tan\pars{\pi\nu}}
_{\ \nu\ =\ 0} =
{\pi \over 8}\,\partiald[2]{}{\nu}
\pars{\pi\nu^{2}}_{\ \nu\ =\ 0}
\\[5mm] = &\
\bbx{\pi^{2} \over 4}\\ &
\end{align}
