Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?

$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$

I know that 
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$

$$
{\cal I}
\equiv
\int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\,\dd x
=
\int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x
\quad\imp\quad
{\cal I}
\equiv
{1 \over 2}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x
$$

Let's consider the integral
$$
{\cal W} \equiv \int_{C}{\ln^{2}\pars{z} \over z^{2} - 1}\,\dd z
=
0
$$

Then
$\pars{~\mbox{with}\ z_{\pm} = x \pm \ic 0^{+}\ \mbox{and}\
 z_{\pm}^{2} = x^{2} \pm \ic\sgn\pars{x}0^{+}~}$
\begin{align}
&\int_{-\infty}^{0}{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over z_{+}^{2} - 1}\,\dd x
+
\int_{0}^{-\infty}{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over z_{-}^{2} - 1}\,\dd x
\\[3mm]&=
\sum_{\sigma = \pm 1}\sigma\int_{-\infty}^{0}
{\ln^{2}\pars{-x} + 2\ic\pi\sigma\ln\pars{-x} - \pi^{2} \over x^{2} - 1 - \sigma\,\ic 0^{+}}\,\dd x
\\[3mm]&=
\sum_{\sigma = \pm 1}\sigma\,\braces{{\cal P}\int_{0}^{\infty}
{\ln^{2}\pars{x} + 2\ic\pi\sigma\ln\pars{x} - \pi^{2} 
\over
x^{2} - 1
}\,\dd x
+
\int_{0}^{\infty}
{\bracks{\ln^{2}\pars{x} + 2\ic\pi\sigma\ln\pars{x} - \pi^{2}}
\bracks{\ic\pi\sigma\delta\pars{x^{2} - 1}} 
}\,\dd x}
\\[3mm]&=
4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x
+
\pars{-\pi^{2}}2\pars{\ic\pi \over 2}
=
0
\quad\imp\quad
\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x = {\pi^{2} \over 4}
\end{align}
$${\large%
{\cal I} = \int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\,\dd x = {\pi^{2} \over 8}}
$$
A: We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$
We will generalize by introducing parameter $\alpha$ such that
$$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$
And we have $I(0)=0$
Then
$$I'(\alpha )=-\int_0^1\frac{1}{(1-\alpha x)(x-2)}\,\mathrm dx=\frac{1}{2\alpha -1}\left[\ln\left(\frac{x-2}{1-\alpha x}\right)\right]_0^1=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
And we have
$$I'(\alpha )=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+c$$
$$I(0)=\frac{1}{2}\text{Li}_2(-1)+c=0\implies c=-\frac{1}{2}\text{Li}_2(-1)=\frac{\pi^2}{24}$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+\frac{\pi^2}{24}$$
$$\begin{align}I(1)&=\frac{1}{2}\text{Li}_2(1)+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{12}+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{8}\\
\end{align}$$

$$I(1)=\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\frac{\pi^2}{8}$$

A: Here is another way to solve. Using the substitute $u=x^2$ gives
\begin{eqnarray}
\int_0^1\frac{\ln x}{x^2-1}dx&=&-\frac{1}{4}\int_0^1\frac{\ln u}{(1-u)\sqrt u}dx\\
&=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\partial}{\partial \mu}\int_0^1(1-u)^{\varepsilon-1}u^{\mu-1}du\\
&=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\partial}{\partial \mu}B(\varepsilon,\mu)\\
&=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}B(\varepsilon,\mu)(\psi(\mu)-\psi(\mu+\varepsilon))\\
&=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\Gamma(\varepsilon+\mu)}{\Gamma(\varepsilon)\Gamma(\mu)}(\psi(\mu)-\psi(\mu+\varepsilon))\\
&=&\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\psi(\mu+\varepsilon)-\psi(\mu)}{\varepsilon}\frac{\varepsilon}{\Gamma(\varepsilon)}\frac{\Gamma(\varepsilon+\mu)}{\Gamma(\mu)}\\
&=&\frac{1}4\psi'(\frac{1}{2})\\
&=&\frac{\pi^2}{8}.
\end{eqnarray}
Here we used
$$ \Gamma(\varepsilon)\approx\varepsilon \text{ for small }\varepsilon>0. $$
A: Write the integrand as a sum of fractions and use the polylogarithm function $\mathrm{Li}_2:$
$$f(x):=\int\frac{\ln(x)}{x^2-1}dx=\int\frac{1}{2}\left(\frac{\ln(x)}{x-1}- \frac{\ln(x)}{x+1}\right) dx \\
=\frac{1}{2} \int \frac{\ln(x) dx}{x-1}- \frac{1}{2} \int \frac{\ln(x)dx }{x+1}
=-\frac{1}{2}\mathrm{Li}_2(1-x) -\frac{1}{2} \left(\mathrm{Li}_2(-x) + \ln(x)\ln(x+1)\right)$$
Since $\mathrm{Li}_2(0)=0$  and $\ln(x)\ln(x+1)$ vanishes at $x=0$ and $x=1$, we have 
$$f(0) = -\frac{1}{2}\mathrm{Li}_2(1)= -\frac{\pi^2}{12}$$ 
and 
$$f(1) = -\frac{1}{2}\mathrm{Li}_2(-1)= \frac{\pi^2}{24}$$
and the value of the integral is 
$$
\int_0^1\frac{\ln(x)}{x^2-1}dx=\frac{\pi^2}{24}+\frac{\pi^2}{12} = \frac{\pi^2}{8}
$$
A: This comes from my answer here.  Let 
$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$
Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value.  We evaluate this integral by once again appealing to the residue theorem, but this time, we consider 
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1}$$
where $C'$ is a keyhole contour with respect to the positive real axis.  By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C'$ increases without bound, we get
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi  \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$
This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C'$.  As the only pole is at $z=-1$, we see that
$$\begin{align}\oint_{C'} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi  \frac{\pi^2}{2}\end{align}$$ 
Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$.  The Cauchy principal value is zero:
$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon}  \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$
The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:
$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$
so that
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$
Therefore
$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{4}$$
But the sought-after integral is
$$\int_0^{1} dx \frac{\log{x}}{x^2-1} = \frac12 I = \frac{\pi^2}{8}$$
A: $$\begin{align}\int_0^1\frac{\ln(x)}{x^2-1}dx&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)\\
&=\frac{1}{2}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{1}{2}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du
=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\ln\left(\frac{1}{u^2}\right)du\\
&=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du\\
&=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du\\
&=\frac{1}{4}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\right)dv\\
&=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\tan^{-1}(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{4}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv\\
&=\frac{\pi}{8}\int_0^{\infty}\frac{2w}{w(1+w^2)}dw=\frac{\pi}{8}\times2\left[\tan^{-1}(w)\right]_0^{\infty}=\frac{\pi}{8}\times2\times\frac{\pi}{2}=\frac{\pi^2}{8}.\end{align}$$
A: Consider the integral:
$$I(m)=\int _{0}^{1}\!{\frac {  \ln\left( x \right)  ^{m-1}}{
{x}^{2}-1}}{dx} \quad:\quad \mathfrak{R}(m)>1 $$
and the substitution $x=e^{-u}$:
$$\begin{aligned}
\int _{0}^{1}\!{\frac {  \ln\left( x \right)  ^{m-1}}{
{x}^{2}-1}}{dx}=& \left( -1 \right) ^{m-1}\int _{0}^{\infty }\!{\frac {
{u}^{m-1}{{\rm e}^{-u}}}{-1+{{\rm e}^{-2\,u}}}}{du}\\
=&\left( -1 \right) ^{m-1}
\int _{0}^{\infty }\!-{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}+{\frac {{u
}^{m-1}}{-1+{{\rm e}^{2\,u}}}}{du}\\
=&\left( -1 \right) ^{m-1}\left( 
1- \dfrac{1}{2^m} \right) 
\int _{0}^{\infty }\!{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}{du}\\
=&\left( -1 \right) ^{m}\left( 
1- \dfrac{1}{2^m} \right) 
\Gamma  \left( m \right) \zeta  \left( m \right) 
\end{aligned}$$
where we have used Riemann's integral representation of the zeta function and we also made the substitution $u\rightarrow\frac{u}{2}$ in the second term of the second line to pass to line three (having noted that convergence of both terms individually is assured by comparison with Riemanns integral). It follows from $\Gamma(2)=1, \zeta(2)=\frac{\pi^2}{6}$ that: $$I(2)=\frac{\pi^2}{8}$$
A: Here is a Feynman-integral-trick approach, with rigorous derivation. Consider for $a>0$
$$
I(a):=\int_{x=0}^1\frac{\ln(1+a^2x^2)-\ln(1+a^2)}{x^2-1}\,dx\tag1
$$
which, by partial fraction decomposition of the integrand, equals
$$
\int_{x=0}^1\left(\int_{t=0}^a\frac{2t}{(1+t^2x^2)(1+t^2)}\,dt\right)\,dx.
$$
Interchange order of integration to find
$$
I(a)=\int_{t=0}^a\frac{2}{1+t^2}\left(\int_{x=0}^1\frac{t\,dx}{1+t^2x^2}\right)\,dt
=\int_{t=0}^a\frac{2\arctan(t)\,dt}{1+t^2}
=\int_{u=0}^{\arctan a}2u\,du=(\arctan a)^2.
$$
The integrand in $(1)$ is non-negative and increases to $\displaystyle\frac{\ln(x^2)}{x^2-1}$ as $a\to\infty$. So by monotone convergence, it is legal to interchange the limit and integration operations, obtaining
$$
\int_{x=0}^1\frac{\ln (x^2)}{x^2-1}\,dx=\lim_{a\to\infty}I(a)=\left(\frac\pi2\right)^2.$$
