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\begin{align}
&\int_{-1}^{1}{\dd x \over x}\,\root{1 + x \over 1 - x}\
{2x\pars{x - 1} + \ln\pars{\bracks{1 - x}\bracks{1 + x}^{3}]} \over x}
\\[5mm] =\ &\
\overbrace{2\int_{-1}^{1}\!\!\root{1 + x \over 1 - x}\,\dd x}^{\ds{2\pi}}\,\,\, +\
\int_{-1}^{1}\!\!\root{1 + x \over 1 - x}\
{\ln\pars{1 - x} + x \over x^{2}}\,\dd x + 3
\int_{-1}^{1}\!\!\root{1 + x \over 1 - x}\
{\ln\pars{1 + x} - x \over x^{2}}\,\dd x
\\[1cm] & =
2\pi + 2\int_{-1}^{1}\root{1 + x \over 1 - x}\
{\bracks{\ln\pars{1 - x} + x} + \bracks{\ln\pars{1 + x} - x}\over x^{2}}\,\dd x
\\[5mm] & \phantom{=\ 2\pi\ } +
\int_{-1}^{1}\root{1 + x \over 1 - x}\
{\bracks{-\ln\pars{1 - x} - x} + \bracks{\ln\pars{1 + x} - x} \over x^{2}}
\,\dd x
\\[1cm] & =
2\pi + 2\int_{-1}^{1}\root{1 + x \over 1 - x}\
{\ln\pars{1 - x^{2}} \over x^{2}}\,\dd x +
\int_{-1}^{1}\root{1 + x \over 1 - x}\bracks{\ln\pars{1 + x \over 1 - x} - 2x}
\,{\dd x \over x^{2}}
\\[5mm] & =
2\pi + 4\int_{0}^{1}
{\ln\pars{1 - x^{2}} \over x^{2}\root{1 - x^{2}}}\,\dd x +
\int_{0}^{1}\pars{\root{1 + x \over 1 - x} - \root{1 - x \over 1 + x}}
\bracks{\ln\pars{1 + x \over 1 - x} - 2x}\,{\dd x \over x^{2}}
\end{align}
En la primera integral me voy a fijar $\ds{x^{2} \mapsto x}$, mientras que en el segundo
Me voy a fijar $\ds{\root{1 + x \over 1 - x} \mapsto x}$. Entonces, la expresión anterior se reduce a:
\begin{align}
&2\pi + 2\ \underbrace{\int_{0}^{1}
{x^{-3/2}\ln\pars{1 - x} \over \root{1 - x}}\,\dd x}_{\ds{-2\pi}}\ -\
8\ \underbrace{\int_{1}^{\infty}{\ln\pars{x} \over 1 - x^{2}}\,\dd x}
_{\ds{-\,{\pi^{2} \over 8}}} -
8\ \underbrace{\int_{1}^{\infty}{\dd x \over 1 + x^{2}}}_{\ds{\pi \over 4}} =
\bbx{\ds{\pi^{2} - 4\pi}}
\end{align}
La primera y de segunda integral puede ser sencillo evaluado por la explotación de su relación con la Función Beta.
