Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $ 
Evaluate : 
$$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$  

Introduction :  I have a friend on another math platform who proposed a summation question and he has a good reputation of posting legitimate questions. I worked it out to another equivalent form i.e, the above integral. Here's my work :- 

We start with  $\displaystyle \sum_{n=0}^\infty L_{2n+1} x^n = \dfrac{x+1}{x^2-3x + 1}  $. Replacing $x$ with $x^2$ , we get
$$ \sum_{n=0}^\infty L_{2n+1} x^{2n} = \dfrac{x^2+1}{x^4-3x^2+1} $$
Integrate:
$$ \sum_{n=0}^\infty \dfrac{L_{2n+1} x^{2n+1}}{2n+1} = \underbrace{\int \dfrac{x^2+1}{x^4-3x^2+1} \,\mathrm{d}x}_{:= I} $$
Then, $\displaystyle I = \int \dfrac{ 1+(1/x^2)}{x^2 + 1/x^2 - 3} \, \mathrm{d}x $. Let $t = x - \dfrac1x  \Rightarrow \left( 1 + \dfrac1{x^2} \right) \, \mathrm{d}x = \mathrm{d}t $.
Then $\displaystyle I = \int \dfrac{\mathrm{d}t}{t^2-1} = \dfrac12 \log \left | \dfrac{t-1}{t+1} \right | = \dfrac12 \log \left | \dfrac{x^2-x-1}{x^2+x-1} \right | $.
$$ \begin{eqnarray}
S & := & \sum_{n=0}^\infty \dfrac{ L_{2n+1}}{(2n+1)^2 \binom{2n}n } = \int_0^1 \sum_{n=0}^\infty \dfrac{ L_{2n+1}}{2n+1} (x-x^2)^n \, \mathrm{d}x  \qquad \left(\text{ Because }\dfrac1{(2n+1) \binom{2n}n} = \operatorname{B}(n+1,n+1) = \int_{0}^{1}  x^n(1-x)^n \mathrm{d}x\right) \\
&=& \int_0^1 \dfrac1{2\sqrt{x-x^2} } \log \left | \dfrac{x -x^2 - \sqrt{x-x^2} - 1}{x -x^2 + \sqrt{x-x^2} - 1} \right | \, \mathrm{d}x \\
&=& \int_0^1 f(x)\, \mathrm{d}x 
\end{eqnarray} $$
Note that $f(1-x) = f(x) $, so $\displaystyle S =2  \int_0^{1/2} f(x) \, \mathrm{d}x = 2 \int_0^{1/2} f\left( \dfrac12 - x\right) \, \mathrm{d}x $, and so
$$ S = \int_0^{1/2} \dfrac1{\sqrt{\frac14 - x^2}} \log \left |\dfrac{a^2-a-1}{a^2+a-1} \right| \, \mathrm{d}x $$
where $a = \sqrt{\dfrac14 - x^2} $.
Substitute $x = \dfrac12 \cos (\theta) $ and simplify:
$$ S = \int_0^{\pi/2} \log \left | \dfrac{ \cos^2 -2 \cos \theta - 4}{\cos^2 + 2\cos\theta - 4} \right | \, \mathrm{d}x = \int_0^1 \log \left ( \dfrac{x^2-2x-4}{x^2+2x-4} \right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} \\ \vdots $$    

Closed Form : Recently, the same question was posted on M.S.E. albeit in a different form by another friend of mine here. That integral is obtained from this by applying Integration By Parts. Mr. Jack D'Aurizio also gave a closed form in terms of Imaginary part of Dilogarithms, specifically, $$I = -2 \ \Im \left[\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right] \right]$$ However, there is a more elementary closed form that exists for the question (as evident from the original question) in terms of natural logarithm and Catalan's constant.  
All solutions are greatly appreciated.
 A: 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:
$$ K=\sum_{n\geq 0}\frac{(-1)^{n}}{(2n+1)^2}=2\int_{0}^{\pi/2}\frac{\theta\,d\theta}{\sin\theta}.\tag{7}$$
