Solving integral: $\int_0^1\ln^2{\left(\frac{1+x}{1-x}\right)} dx$ I want to show that the solution of a BVP: $\ u(x) = \ln{\frac{1+x}{1-x}}$ is in $L^2(0,1)$, so I need to show that the integral
$$\int\limits_0^1\ln^2|\frac{x+1}{x-1}|dx < \infty$$
Just looking at the function, however, it's not even defined in the interval $[0,1]$, right? So can this not actually be a solution to the BVP? Even more generally about the integral itself, does that make it just $0$? And if not, can someone explain how an integral of a function can be defined in a region when the function itself is not?
Also, I know that there exist integral calculators online and have looked this up, specifically using:
https://www.integral-calculator.com/
If you put in the given integral it says the integral could not be found and then gives a complex number approximation. What should be made of this? I have taken a complex variables course but don't really see how you could solve this using Residue Calculus since we don't have any symmetries.
EDIT: Note I had originally forgot the absolute value signs - my mistake.

Substitute $\frac{1-x}{1+x}\to x $
\begin{align}
\int_0^1\ln^2{\frac{1+x}{1-x}} dx 
= &\int_0^1\frac{2\ln^2{x}}{(1+x)^2}dx= \int_0^1{\ln^2{x}}\>d\left(\frac {2x}{1+x}\right)\\
=& -4\int_0^1\frac{\ln x}{1+x}dx=-4\cdot (-\frac{\pi^2}{12})=\frac{\pi^2}3
\end{align}
$ \int_0^1\frac{\ln x}{1+x}dx =-\int_0^1\frac{\ln (1+x)}xdx =-\frac{\pi^2}{12}$
