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.
 A: $$
\begin{align}
\int_0^1\log\left(\frac{1+x}{1-x}\right)^2\mathrm{d}x
&=\int_1^\infty\log(u)^2\frac{2\,\mathrm{d}u}{(u+1)^2}\tag1\\
&=4\int_1^\infty\frac1{u+1}\frac{\log(u)}u\,\mathrm{d}u\tag2\\
&=-4\int_0^1\frac{\log(u)}{u+1}\,\mathrm{d}u\tag3\\
&=-4\sum_{n=0}^\infty\int_0^1(-1)^nu^n\log(u)\,\mathrm{d}u\tag4\\
&=4\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}\tag5\\
&=4\cdot\frac{\pi^2}{12}\tag6\\[3pt]
&=\frac{\pi^2}3\tag7
\end{align}
$$
Explanation:
$(1)$: $u=\frac{1+x}{1-x}\implies x=\frac{u-1}{u+1}$ and $\mathrm{d}x=\frac{2\,\mathrm{d}u}{(u+1)^2}$
$(2)$: integrate by parts
$(3)$: substitute $u\mapsto1/u$
$(4)$: apply the Taylor series for $\frac1{1+u}$
$(5)$: $\int_0^1u^n\log(u)\,\mathrm{d}u=-\frac1{(n+1)^2}$
$(6)$: $\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\sum\limits_{n=1}^\infty\frac1{n^2}-2\sum\limits_{n=1}^\infty\frac1{(2n)^2}=\frac{\pi^2}6-\frac{\pi^2}{12}$
$(7)$: simplify
A: 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}$
A: Actually, this integral is very straightforward to evaluate using Cauchy's Theorem in the complex plane.
Consider the complex integral
$$\oint_C dz \, \log^3{\left (\frac{z+1}{z-1} \right )} $$
where $C$ is the following contour in the complex plane:

The outer arc of $C$ has a radius $R$ and the small circular pieces around the branch points at $z=\pm 1$ have radius $\epsilon$.  The contour integral is then equal to
$$e^{i \pi} \int_R^{1+\epsilon} dx \, \log^3{\left (\frac{x-1}{x+1} \right )} + i \epsilon \int_{\pi}^0 dx \, e^{i \phi} \, \log^3{\left (\frac{-1+\epsilon e^{i \phi}+1}{-1+\epsilon e^{i \phi}-1} \right )}  \\ + \int_{-1+\epsilon}^{1-\epsilon} dx \, \left [\log{\left (\frac{1+x}{1-x} \right )} -i \pi\right ]^3 + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, \log^3{\left (\frac{1+\epsilon e^{i \phi}+1}{1+\epsilon e^{i \phi}-1} \right )}  \\ - \int_{-1+\epsilon}^{1-\epsilon} dx \, \left [\log{\left (\frac{1+x}{1-x} \right )} +i \pi \right ]^3 + i \epsilon \int_{2 \pi}^{\pi} dx \, e^{i \phi} \, \log^3{\left (\frac{-1+\epsilon e^{i \phi}+1}{-1+\epsilon e^{i \phi}-1} \right )}  \\ - e^{-i \pi} \int_R^{1+\epsilon} dx \, \log^3{\left (\frac{x-1}{x+1} \right )} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \, \log^3{\left (\frac{R e^{i \theta}+1}{R e^{i \theta}-1} \right )} $$
As $\epsilon \to 0$ and $R \to \infty$, all integrals vanish or cancel except the third and fifth; the contour integral is then equal to
$$\int_{-1}^1 dx \, \left \{\left [\log{\left (\frac{1+x}{1-x} \right )} -i \pi\right ]^3 -\left [\log{\left (\frac{1+x}{1-x} \right )} +i \pi\right ]^3  \right \} $$
By Cauchy's Theorem, the contour integral is equal to zero.  Expanding the integrand, we immediately determine the value of the integral without further computation, because the cube and linear powers of the log cancel.  That is, we are left with the log squared term sought, and a constant term that is trivially dealt with.  The result is, using symmetry of an even function:
$$\int_0^1 dx \, \log^2{\left (\frac{1+x}{1-x} \right )} = \frac{\pi^2}{3}$$
A: A handy substitution when our limits are from $x=0$ and $x=1$ is $$t=\frac {1-x}{1+x}\qquad\implies\qquad\mathrm dx=-\frac {2\,\mathrm dt}{(1+t)^2}$$
The integral becomes
$$\begin{align*}\mathfrak{I} & =\int\limits_0^1\mathrm dx\,\log^2\left(\frac {1-x}{1+x}\right)\\ & =2\int\limits_0^1\mathrm dt\,\frac {\log^2t}{(1+t)^2}\end{align*}$$
Now recall the geometric series formula
$$\sum\limits_{n\geq1}x^{n-1}=\frac 1{1-x}$$
Differentiating with respect to $x$ once gives
$$\sum\limits_{n\geq1}nx^{n-1}=\frac 1{(1-x)^2}$$
Replacing the integrand with our modified geometric sequence, then
$$\begin{align*}\mathfrak{I} & =2\sum\limits_{n\geq1}n(-1)^{n-1}\int\limits_0^1\mathrm dt\, t^{n-1}\log^2t\\ & \stackrel{\text{IBP}}{=}4\sum\limits_{n\geq1}\frac {(-1)^{n-1}}{n^2}\end{align*}$$
If you're familiar with Basel's problem, then it's easy to see through some infinite sum manipulation that
$$\begin{align*}\sum\limits_{n\geq1}\frac {(-1)^{n-1}}{n^2} & =\sum\limits_{n\geq1}\frac 1{n^2}-2\sum\limits_{n\geq1}\frac 1{(2n)^2}\\ & =\frac {\pi^2}{12}\end{align*}$$
Hence
$$\int\limits_0^1\mathrm dx\,\log^2\left(\frac {1-x}{1+x}\right)=4\left(\frac {\pi^2}{12}\right)=\frac {\pi^2}{3}$$
A: Do you know that :
$$f(x):=\operatorname{ln}\frac{1+x}{1-x}=2\operatorname{arctanh}(x) \ ? \ \tag{1}$$
Remark: $\operatorname{arctanh}$ is the same as $\operatorname{tanh}^{-1}$.
Besides, Wolfram Alpha gives
$$\int_0^1 \operatorname{arctanh}(x)^2 dx= \pi^2/12$$
Therefore your result is $\pi^2/48 < \infty$
Wolfram Alpha also gives an (awful) expression for a primitive function of $f$.
