Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$

$$\begin{align}\int_0^1\frac{\ln(x)}{x^2-1}dx&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)\\ &=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)\\ &=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)\\ &=\frac{1}{2}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{1}{2}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du =\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\ln\left(\frac{1}{u^2}\right)du\\ &=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du\\ &=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du\\ &=\frac{1}{4}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\right)dv\\ &=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\tan^{-1}(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{4}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv\\ &=\frac{\pi}{8}\int_0^{\infty}\frac{2w}{w(1+w^2)}dw=\frac{\pi}{8}\times2\left[\tan^{-1}(w)\right]_0^{\infty}=\frac{\pi}{8}\times2\times\frac{\pi}{2}=\frac{\pi^2}{8}.\end{align}$$

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: $$I(2)=\frac{\pi^2}{8}$$

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

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} $${\large% {\cal I} = \int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\,\dd x = {\pi^{2} \over 8}} $$