How to prove $\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1+x\over 1-x}\cdot{2x(x-1)+\ln((1-x)(1+x)^3)\over x}=\pi^2-4\pi?$ Motivated by this interesting question and ideas came from lemma 2.
Then we have this integral:

$$\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1+x\over 1-x}\cdot{2x(x-1)+\ln((1-x)(1+x)^3)\over x}=\pi^2-4\pi.\tag1$$

Here is my try:
Split out $(1)$ then we have
$$2\int_{-1}^{1}{\mathrm dx\over x}\cdot\sqrt{1-x^2}+\int_{-1}^{1}{\mathrm dx\over x^2}\cdot\sqrt{1+x\over 1-x}\cdot\ln((1-x)(1+x)^3)=I_1+I_2.\tag2$$
For $I_1$, let $x=\sin u$ then we have
$$I_1=\int_{-\pi/2}^{\pi/2}\mathrm du{\cos^2(u)\over \sin(u)},\tag3$$
as for $I_2$ seems too complicate to make an attempt. Probably easy but I can't see it.
How to prove (1) and (2)?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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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}
In the first integral I'll set $\ds{x^{2} \mapsto x}$ while in the second one
I'll set $\ds{\root{1 + x \over 1 - x} \mapsto x}$. Then, the above expression is reduced to:
\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}
The first and de second integral can be straightforward evaluated by exploiting its relation to the Beta Function.
