<s>[INST] 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 ?
 [/INST] Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
when the transformation $x = \tanh(t)$ is made. The resulting integral is given by
\begin{align}
I &= \int_{0}^{\infty} \ln\left( \frac{1+\tanh(t)}{1-\tanh(t)}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\
&= \int_{0}^{\infty} \ln\left( \frac{e^{t}}{e^{-t}}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\
&= 2 \int_{0}^{\infty} \frac{t \cosh^{2}(t)}{\sinh(t)} \ dt \\
&= 2 \left( \frac{\pi^{2}}{4} \right)
\end{align}
which yields
\begin{align}
\int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx = \frac{\pi^{2}}{2}.
\end{align}
</s>