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 ?

$\displaystyle J=\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$
Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,
$\displaystyle J=-4\int^1_0 \dfrac{\log(x)}{1-x^2}dx=-4\int_0^1 \left( \log x\times\sum_{n=0}^{+\infty}x^{2n}\right) dx=-4\sum_{n=0}^{+\infty}\left(\int_0^1 x^{2n}\log x dx\right)$
$\displaystyle J=4\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=4\left(\zeta(2)-\sum_{n=1}^{+\infty}\dfrac{1}{(2n)^2}\right)=4\times\left(1-\dfrac{1}{4}\right)\times\zeta(2)=4\times\dfrac{3}{4}\times\dfrac{\pi^2}{6}=\dfrac{\pi^2}{2}$
Note that:
0) For $s>1$, $\displaystyle \zeta(s)=\sum_{n=1}^{+\infty} \dfrac{1}{n^s}$
1) $\zeta(2)=\dfrac{\pi^2}{6}$
2) For $n\geq 0$, $\displaystyle \int_0^1  x^n\log x dx=\left[\dfrac{x^{n+1}}{n+1}\times\log x\right]_0^1-\int_0^1  \left(\dfrac{x^{n+1}}{n+1}\times\dfrac{1}{x}\right)dx=-\left[\dfrac{x^n}{n+1}\right]_0^1=-\dfrac{1}{(n+1)^2}$
