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!

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 $\displaystyle2\left(\frac{\pi^2}{6}-\frac{\pi^2}{24}\right)=\frac{\pi^2}{4}$.
