Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$  
The first idea was to use the integration by parts because  
$$
\int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,}
=\arcsin\left(\,2x - 1\,\right) + C
$$   
but what would be the next step is not clear. Another idea would be expand
$\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.   
So, what are the other options?

Let $x = \sin(t)^2$ and $s = 2t$, we have
$$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}}
= \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{
\sqrt{\sin(t)^2(1-\sin(t)^2)}}\\
= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt
= \int_0^{\pi}\log\left(\frac{3 + \frac{1-\cos s}{2}}{3-\frac{1-\cos s}{2}}\right) ds\\
= \int_0^{\pi}\left(\log(7-\cos s) - \log(5+\cos s)\right) ds
$$
Notice for any $a > 1$, we have
$$\frac{1}{\pi}\int_0^\pi \log(a \pm \cos s)ds = \cosh^{-1}(a) = \log\left(\frac{a + \sqrt{a^2-1}}{2}\right)\tag{*1}$$
The integral we desired is simply 
$$\pi\left( \cosh^{-1}(7) - \cosh^{-1}(5)\right)
= \pi \log\left(\frac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\approx 1.072804016182156$$
I'm sure the identity in $(*1)$ has a name but I can't remember what it is. Let us 
prove it!
Notice for any $b > 1$, the function $\log(b+z)$ is analytic over and inside the unit circle $S^1$ in $\mathbb{C}$.
By Residue theorem, we have
$$\frac{1}{2\pi i}\int_{S^1} \log(b + z) \frac{dz}{z} = \log(b)$$
If one parametrize the unit circle by $z = e^{i\theta}$, we get
$$\frac{1}{2\pi}\int_0^{2\pi} \log(b + e^{i\theta}) d\theta = \log(b)$$
Take the real part on both sides, this leads to
$$\begin{align}
&\frac{1}{2\pi}\int_0^{2\pi} \log(b^2 + 1 + 2b\cos\theta) d\theta = \log(b^2)\\
\iff &
\frac{1}{2\pi}\int_0^{2\pi} \log\left(\frac{b+b^{-1}}{2} + \cos\theta\right)d\theta = \log\left(\frac{b}{2}\right)\end{align}$$
Substitute $\displaystyle\;\frac{b+b^{-1}}{2}\;$ by $a$, we have
$\displaystyle\;\frac{b-b^{-1}}{2} = \sqrt{a^2-1}$ and it is clear $(*1)$ follows.
