A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following 
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.
Any other ideas ?

Let $-1\le a \le 1$ and:
\begin{align*}
  I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx \tag 1\\
  \frac{\partial}{\partial a}I(a) &= \int_{0}^{1} \, \frac{1}{(1+a\, x)\sqrt{1-x^2}} + \frac{1}{(1-a\, x)\sqrt{1-x^2}} \, dx\\
  &= \frac{1}{\sqrt{1-a^2}}\, \left(\arcsin\left(\frac{x+a}{1+a\, x}\right)+\arcsin\left(\frac{x-a}{1-a\, x}\right) \right) \Big|_0^1\\
  &= \frac{\pi}{\sqrt{1-a^2}}\\
  \therefore I(a) &= \pi\, \arcsin{a} + C \tag 2\\
\end{align*}
Putting $a=0$, in $(1)$ and $(2)$, we see that $C=0$
Hence, 
\begin{align*}
    I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx = \pi\, \arcsin{a}
\end{align*}
and for this problem 
$$I(1)=\frac{\pi^2}{2}$$
