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?
 A: Split the integral into two forms by expanding the logarithm function
$$
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx
$$
Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have
$$
2\int_{0}^1\frac{\ln(3+t^2)}{\sqrt{1-t^2}}\ dt-2\int_{0}^1\frac{\ln(3-t^2)}{\sqrt{1-t^2}}\ dt
$$
Let $t=\sin\theta\ \rightarrow\ dt=\cos\theta\ d\theta$, we have
$$
2\int_{0}^{\pi/2}\ln(3+\sin^2\theta)\ d\theta-2\int_{0}^{\pi/2}\ln(3-\sin^2\theta)\ d\theta
$$
Using identity $\sin^2\theta=\dfrac12(1-\cos2\theta)$ and setting $y=2\theta$, we have
$$
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^{\pi}\ln\left(7-\cos y\right)\ dy-\int_{0}^{\pi}\ln\left(5+\cos y\right)\ dy
$$
We will use Feynman's way to evaluate integral above. Consider
$$
I(k)=\int_{0}^{\large\pi}\ln\left(k\pm\cos y\right)\ dy
$$
then
\begin{align}
I'(k)&=\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\ dy
\end{align}
Using formula
$$
\int_0^\pi\frac{1}{a^2+b^2-2ab\cos x}dx=\frac{\pi}{a^2-b^2}
$$
Setting $b=\pm\dfrac{1}{2a}$, we have
$$
\int_0^\pi\frac{1}{a^2+\frac{1}{4a^2}\pm\cos x}dx=\frac{4\pi a^2}{4a^4-1}
$$
Clearly $k=a^2+\dfrac{1}{4a^2}\ \rightarrow\ a^2=\dfrac{k+\sqrt{k^2-1}}{2}$ and $dk=\dfrac{4a^4-1}{2a^3}da$, then
\begin{align}
I(k)&=\int\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\,dy\,dk\\
&=\int\frac{4\pi a^2}{4a^4-1}\cdot\dfrac{4a^4-1}{2a^3}da\\
&=2\pi\int\frac{1}{a}\,da\\
&=2\pi\ln(a)+C\\
&=\pi\ln(a^2)+C\\
&=\pi\ln\left(\dfrac{k+\sqrt{k^2-1}}{2}\right)+C\\
\end{align}
Finally
\begin{align}
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx&=I(7)-I(5)\\
&=\pi\ln\left(\dfrac{7+\sqrt{7^2-1}}{5+\sqrt{5^2-1}}\right)\\
&=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\
\end{align}
Yeayyy, I'm done! (>‿◠)✌
A: 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.
A: 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:

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

