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 ?

Here is a solution, which I discovered:
$$I=\int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{x(\sqrt{1-x^2})}dx=\int_{0}^{1} \frac{2\tanh^{-1}(x)}{x(\sqrt{1-x^2})}dx.$$
Observe $$I=\int_{0}^{1} \int_{0}^{1} \frac{2}{(1-x^2y^2)\sqrt{1-x^2}}  dy dx$$ by using the fact that $$\int_{0}^{1}\frac{1}{1-x^2y^2}dy=\frac{\tanh^{-1}(x)}{x}.$$ Now to evaluate $I,$ make the change of variables $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(u)}$$ which has Jacobian $$\frac{\partial(x,y)}{\partial(u,v)}=1-x^2y^2.$$ The region of integration becomes the open triangle formed by the inequalities $0<u+v<\frac{\pi}{2}$ with $u,v>0.$ We now get 
$$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2}{\sqrt{1-\left(\frac{\sin(u)}{\cos(v)}\right)^2}}dvdu=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2\cos(v)}{\sqrt{(\cos(v))^2-(\sin(u))^2}}dvdu.$$
We can rewrite $I$ as $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2\cos(v)}{\sqrt{1-(\sin(v))^2-(\sin(u))^2}}dvdu.$$ Letting $t=\sin(v),dt=\cos(v)dv$ we get $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos(u)}\frac{2}{\sqrt{1-t^2-(\sin(u))^2}}dtdu$$ Simplifying a bit more, we get $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos(u)}\frac{2}{\sqrt{(\cos(u))^2-t^2}}dtdu.$$ Using the fact $$\int_{0}^{\cos(u)}\frac{1}{\sqrt{(\cos(u))^2-t^2}}dt=\lim_{t \rightarrow \cos(u)}\sin^{-1}\left(\frac{t}{\cos(u)}\right)=\frac{\pi}{2},$$ we see $$I= \int_{0}^{\frac{\pi}{2}} 2\left(\frac{\pi}{2}\right)du=\frac{\pi^2}{2}.$$
