<s>[INST] 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?
 [/INST] We have:

$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$

This happens because:
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$
$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=\frac{1}{2}\int_{-\pi}^{\pi}\log\left(\frac{7+\cos\theta}{2}\right)d\theta=-\pi\log 2+\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta.$$
Now comes an interesting technique - we have:
$$\begin{eqnarray*}\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta &=& \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n}\log\left(7+\cos\frac{k\pi}{n}\right)\\&=&\lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n}\left(7+\cos\frac{k\pi}{n}\right)\end{eqnarray*}$$
but since:
$$z^{2n}-1 = \prod_{k=1}^{2n}\left(z-e^{\frac{\pi i k }{n}}\right)=(z^2-1)\prod_{k=1}^{n-1}\left(z^2+1-2z\cos\frac{k\pi}{n}\right)$$
and the solutions of 
$$\frac{z^2+1}{2z}=-7$$
are $z=\pm 4\sqrt{3}-7$, it follows that:
$$\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta=\pi\log\left(\frac{7}{2}+2\sqrt{3}\right).$$
With the same technique we can prove:

$$\int_{0}^{1}\frac{\log(3-x)}{\sqrt{x(1-x)}}\,dx = \pi\log\left(\frac{5}{4}+\sqrt{\frac{3}{2}}\right),\tag{2}$$

hence we have:

$$\int_{0}^{1}\frac{\log\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\,dx = \pi\log\left((7+4\sqrt{3})(5-2\sqrt{6})\right).\tag{3}$$

</s>