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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)?

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  • $\begingroup$ Your $I_1$ diverges. $\endgroup$ – xpaul Apr 27 '17 at 21:58
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

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