Solve the following definite integral: $\int_{0}^{\infty}\frac{x^2dx}{({1-x^2})^2}$ 
Solve the following integral:
$$\int_{0}^{∞}\frac{x^2dx}{({1-x^2})^2}$$

I know that substituting some trigonometric functions may help. 
But I was not able to solve. Can you give me some clues?
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$\large\mbox{In case the}\ -\ \mbox{switch to}\ +$:

$$
\mu > 0\,,
\quad
\int_{0}^{\infty}{1 \over 1 + \mu x^{2}}\,{\rm d}x
=
{1 \over \sqrt{\mu\,}}\int_{0}^{\infty}{1 \over 1 + x^{2}}\,{\rm d}x
=
{1 \over \sqrt{\mu\,}}\,{\pi \over 2}
$$

Derivate respect of $\ds{\mu}$ in both members:
$$
-\int_{0}^{\infty}{x^{2} \over \left(1 + \mu x^{2}\right)^{2}}\,{\rm d}x
=
-\,{1 \over 2\mu^{3/2}}\,{\pi \over 2}
$$

Set $\mu = 1$:
  $$
\int_{0}^{\infty}{x^{2} \over \left(1 + x^{2}\right)^{2}}\,{\rm d}x
=
{\pi \over 4}
$$

$\large\mbox{With the}\ -\ \mbox{sign, we'll assume it's a 'principal value'}\ \pp
$:
\begin{align}
\pp\int_{0}^{\infty}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x&
=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
\int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x
+\int_{1 + \epsilon}^{\infty}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
\int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x
+\int_{1/\pars{1 + \epsilon}}^{0}{1/x^{2} \over \left(1 - 1/x^{2}\right)^{2}}
\,\pars{-\,{\dd x \over x^{2}}}}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
\int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x
+\int_{0}^{1/\pars{1 + \epsilon}}{x^{2} \over \left(1 - x^{2}\right)^{2}}
\,\dd x}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
2\int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x
+\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}
{x^{2} \over \left(1 - x^{2}\right)^{2}}\,\dd x}
\end{align}
The second integral $\ds{\stackrel{\epsilon\ \to\ 0^{+}}{\to}{1 \over 4}}$
while the first term behaves as $\ds{{1 \over 2\epsilon}}$ when
$\ds{\epsilon \gtrsim 0}$ such that 'even' the "principal value" diverges.
A: Hint:
$\int_0^\infty\dfrac{x^2}{(1-x^2)^2}dx$
$=\int_0^1\dfrac{x^2}{(1-x^2)^2}dx+\int_1^\infty\dfrac{x^2}{(1-x^2)^2}dx$
$=\int_0^\infty\dfrac{\tanh^2x}{(1-\tanh^2x)^2}d(\tanh x)+\int_\infty^0\dfrac{\coth^2x}{(1-\coth^2x)^2}d(\coth x)$
$=\int_0^\infty\dfrac{\text{sech}^2x\tanh^2x}{\text{sech}^4x}dx+\int_0^\infty\dfrac{\text{csch}^2x\coth^2x}{\text{csch}^4x}dx$
$=\int_0^\infty(\sinh^2x+\cosh^2x)~dx$
$=\int_0^\infty\cosh2x~dx$
$=\left[\dfrac{\sinh2x}{2}\right]_0^\infty$
$=+\infty$
$\therefore\int_0^\infty\dfrac{x^2}{(1-x^2)^2}dx$ is divergent.
