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 ?

After this corps has rised from the dead anyway, let me give an additional solution which is based on complex analysis but avoids the use of branch cuts, so it should be viewed as complementary to @RandomVariables approach.
First, perform an subsitution $x\rightarrow\sin(t)$ which brings our integral into the form 
$$
I=\int_0^{\pi/2}\frac{\log(1+\sin(t))-\log(1-\sin(t))}{\sin(t)} dt
$$
Now we introduce a parameter $a$ and differentiate w.r.t. it:
$$
I'(a)=\int_0^{\pi/2}\frac{1}{(a+\sin(t))\sin(t)}-\frac{1}{(a-\sin(t))\sin(t)} dt
$$
Please note the singularity at zero is removeable, so we don't need a Cauchy principle part. We can  reduce the above to
$$
I'(a)=4\int_0^{\pi/2}\frac{1}{2a^2-1+\cos(2t)}dt=\int_0^{2\pi}\frac{1}{2a^2-1+\cos(t)}dt
$$
we now can perform the ususal $e^{it}\rightarrow z$ substitution to map the problem onto a contour integral over the unit circle $C_+$ traveresed counterclockwise:
$$
I'(a)=\frac{2}{i}\int_{C_+}\frac{1}{2z(2a^2-1)+z^2+1}dt
$$
the poles are given by $z_{\pm}=1-2a^2\pm a\sqrt{a^2-1}$. At this stage a little problem appears: For $a < 1$  the poles degenerate and become double poles on our contour which we don't like, so we want to choose $a>1$ and take the limit $a\rightarrow 1_+$ in the end. For this choice of $a$ only $z_+$ lies inside our contour of integration. The corresponding residue is
$$
\text{Res}(z=z_+)=\frac{1}{2 i a\sqrt{a^2-1} }
$$
and therefore

$$
I'(a)=\frac{\pi}{a \sqrt{a^2-1}}
$$

Now what is left is integrating back to $a$. The corresponding integral is elementary and yields

$$
I(a)=\pi\arctan\left(\sqrt{\frac{1}{a^2-1}}\right)+C
$$

The constant of integration can be fixed by the observation that $I(\infty)=0$ yieldung $C=0$
Taking finally the limit $a\rightarrow 1_+$ yields

$$
I=I(1)=\pi \times \arctan(+\infty)=\pi \times \frac{\pi}{2}=\frac{\pi^2}{2}
$$

in accordance with other answers
Remark:
We see that we can avoid branch cuts completly but get instead new complications in choosing the correct limit of $a$ and finding the poles which are inside our contour. In summary both methods work and we just exchange some comperable hard problems at the end of the day, so it's just a matter of taste which one we choose
