Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $
It is really strange how often many problems here on MSE boil down to the same one.
In particular, I am talking about an identity for the squared arcsine function whose consequence is:
$$ \sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4}{\sqrt{x(4-x)}}\arcsin\left(\frac{\sqrt{x}}{2}\right)\, \tag{1}$$
If we take the following series definition of $T$: $$ T = \sum_{n\geq 0}\frac{L_{2n+1}}{(2n+1)\binom{2n}{n}}\tag{2} $$ and recall that $L_{2k+1}=\varphi^{2k+1}+\overline{\varphi}^{2k+1}$, we just have to plug in $x=\varphi^2$ and $x=\overline{\varphi}^2$ in $(1)$ to get the closed form:
$$ T = \color{red}{\frac{3\pi}{5}\sqrt{2+\frac{2}{\sqrt{5}}}-\frac{\pi}{5}\sqrt{2-\frac{2}{\sqrt{5}}}}.\tag{3}$$
If in $(1)$ we replace $x$ with $x^2 z^2$ and integrate over $[0,1]$ with respect to $x$, we get:
$$ \sum_{n\geq 0}\frac{z^{2n+1}}{(2n+1)^2 \binom{2n}{n}} = 2\int_{0}^{z/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=2\int_{0}^{\arcsin(z/2)}\frac{\theta}{\sin\theta}\,d\theta \tag{4}$$
and now the relation with the Clausen function is self-evident.
The antiderivative of $\frac{t}{\sin t}$ is a combination of logarithms and dilogarithms and:
$$ S = 2\int_{0}^{\frac{3\pi}{10}}\frac{\theta\,d\theta}{\sin\theta}-2\int_{0}^{\frac{\pi}{10}}\frac{\theta\,d\theta}{\sin\theta}=2\int_{\pi/10}^{3\pi/10}\frac{\theta\,d\theta}{\sin\theta} \tag{5}$$
simplifies to:
$$ S = \color{red}{K+\frac{\pi}{5}\log(2)}\tag{6} $$
where: $$ K=\sum_{n\geq 0}\frac{(-1)^{n}}{(2n+1)^2}=2\int_{0}^{\pi/2}\frac{\theta\,d\theta}{\sin\theta}.\tag{7}$$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \int_{0}^{1} \ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4}\, {\dd x \over \root{1 -x^{2}}} =\, {\Large ?} $$
$\quad$In this 'answer', I want to summarize the result which involves the Dilogarithm function $\pars{~\mbox{namely,}\ \,\mathrm{Li}_{2}\pars{z}~}$. For this purpose, I found a quite useful result in a previous Vladimir Reshetnikov answer. I'm aware of an interesting issue which has been pointed out by many users along the comments: The relationship between the '$\,\mathrm{Li}_{2}$ result' and the 'Catalan constant result' as it appears in Jack D'Aurizio answer which is somehow still open.
The roots of $\ds{x^{2} - 2x - 4}$ are given by $\ds{2\varphi}$ and $-2\ds{\Phi}$ where $\ds{\varphi}$ and $\ds{\Phi = 1/\varphi}$ are the *G0lden Ratio* $\ds{\root{5} + 1 \over 2}$ and the *Conjugated Golden Ratio* $\ds{\root{5} - 1 \over 2}$, respectively. Similarly, the roots of $\ds{x^{2} + 2x - 4}$ are given by $\ds{-2\varphi}$ and $\ds{2\Phi}$. Namely, \begin{align} &x^{2} - 2x - 4 = \pars{x - 2\varphi} \pars{x + 2\Phi} \\[5mm] = &\ -4\pars{1 - \half\,\Phi x}\pars{1 + \half\,\varphi x} \\[3mm] &\ x^{2} + 2x - 4 = \pars{x + 2\varphi}\pars{x - 2\Phi} \\[5mm] = &\ -4\pars{1 + \half\,\Phi x}\pars{1 - \half\,\varphi x} \end{align} With the sub$\ds{\ldots\ x = \cos\pars{\theta}}$, the above integral is rewritten as: \begin{align} &\color{#f00}{\int_{0}^{1}\ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4} \,{\dd x \over \root{1 -x^{2}}}} \\[3mm] = &\ \bracks{\mathrm{f}\pars{\half\,\Phi} - \mathrm{f}\pars{-\,\half\,\Phi}} - \bracks{% \mathrm{f}\pars{\half\,\varphi} - \mathrm{f}\pars{-\,\half\,\varphi}} \end{align} where $$ \mathrm{f}\pars{t} \equiv \int_{0}^{1}{\ln\pars{1 - tx} \over \root{1 - x^{2}}}\,\dd x = \int_{0}^{\pi/2}\ln\pars{1 - t\cos\pars{\theta}}\,\dd\theta\,,\qquad t \in \pars{-1,1} $$ $\mathrm{f}\pars{t}$ is given by [formula $\pars{4}$ in another Vladimir Reshetnikov answer][1]:
$$ \mathrm{f}\pars{t} = {\pi \over 2}\,\ln\pars{1 + \root{1 - t^{2}} \over 2} - 2\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} $$ Note that \begin{align} &\mathrm{f}\pars{t} - \mathrm{f}\pars{-t} \\[3mm] = &\ 2\,\Im\,\mathrm{Li}_{2}\pars{-\,{1 - \root{1 - t^{2}} \over t}\,\ic} - 2\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} \\[3mm] = &\ -4\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} \end{align}
Since $\ds{\Phi^{2} + \Phi + 1 = \varphi^{2} - \varphi - 1 = 0}$: \begin{align} \left.{1 - \root{1 - t^{2}} \over t}\right\vert_{\ t\ =\ \Phi/2} = {2 - \root{3 + \Phi} \over \Phi}\quad\mbox{and}\quad \left.{1 - \root{1 - t^{2}} \over t}\right\vert_{\ t\ =\ \varphi/2} = {2 - \root{3 - \varphi} \over \varphi} \end{align} Then, \begin{align} &\color{#f00}{\int_{0}^{1}\ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4} \,{\dd x \over \root{1 -x^{2}}}} \\[3mm] = &\ \color{#f00}{% 4\,\Im\,\mathrm{Li}_{2}\pars{{2 - \root{3 - \varphi} \over \varphi}\,\ic} - 4\,\Im\,\mathrm{Li}_{2}\pars{{2 - \root{3 + \Phi} \over \Phi}\,\ic}} \\[3mm] = &\ \color{#f00}{% 4\,\Im\,\mathrm{Li}_{2}\pars{\bracks{-1 + \root{5} - \root{5 - 2\root{5}}}\ic} - 4\,\Im\,\mathrm{Li}_{2}\pars{\bracks{1 + \root{5} - \root{5 + 2\root{5}}}\ic}} \tag{1} \\[3mm] \approx &\ 1.3523870463919131106825397783200\color{#f00}{513308068289818222} \end{align}
This is slightly bigger $\pars{~\sim 1.16\times 10^{-32}~}$ than the 'direct' numerical calculation of the original integral $\pars{~\approx 1.3523870463919131106825397783200\color{#f00}{397004399596207528}~}$. ADDENDA The user @Ishan Singh shows me ( $\mbox{01-jul-2016}$ ) a closed expression: \begin{align} &-\,{\pi \over 5}\, \ln\pars{124 - 55\root{5} + 2\root{7625 - 3410\root{5}}} \\[2mm] &\ + {8 \over 5}\,G \tag{2} \end{align} where $\ds{G}$ is the Catalan Constant. It would be nice to know 'how to travel' between $\pars{1}$ and $\pars{2}$.