<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] 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! (>‿◠)✌
</s>