<s>[INST] How to prove $\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}$? Can anyone suggest the method of computing 
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}\quad ?$$ 

My trial is following 
first set $t =\frac{1-x}{1+x}$ which gives $x=\frac{1-t}{t+1}$ 
Then 
\begin{align}
dx = \frac{2}{(1+t)^2} dt, \quad [x,0,1] \rightarrow [t,1, 0]
\end{align}
[Thanks to  @Alexey Burdin, i found what i do wrong in substitution]
then the integral reduces to
\begin{align} 
\int^{0}_1 \ln(t)\frac{2}{1-t^2} dt 
\end{align}
How one can obtain above integral?
Please post an answer if you know the answer to this integral or the other methods to evaluate above integral. 
Thanks!
 [/INST] Setting the convergence issue aside, I'll compute the integral formally using power series.
Note that
\begin{eqnarray}\frac{\ln\frac{1+x}{1-x}}{x}=\frac{\ln(1+x)-\ln(1-x)}{x}=\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}\end{eqnarray}
Integrating termwise,
\begin{eqnarray}
\int_0^1\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}=2\sum_{n=1}^\infty\frac{1}{(2n-1)^2}
\end{eqnarray}
Using the fact that $\displaystyle\sum_{m=1}^\infty\frac{1}{m^2}=\frac{\pi^2}{6}$ and $\sum_{n=1}^\infty\frac{1}{(2n)^2}=\frac{\pi^2}{24}$, we have the required sum is $\displaystyle2\left(\frac{\pi^2}{6}-\frac{\pi^2}{24}\right)=\frac{\pi^2}{4}$.
</s>