<s>[INST] $\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
My working:
$$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\int_{0}^{1}\frac{\ln x}{x^2-1}dx=-\sum_{r\ge 1}\int_0^{1}x^{2r}\ln x\,dx =\sum_{r\ge 1}\frac {1}{(2r-1)^2}= 
\frac{\pi^2}{8}$$
Is there any other approach?
 [/INST] \begin{align}J&=\int_{0}^{1}\frac{\ln x}{x^2-1}dx\\
&=\frac{1}{4}\int_{0}^{1}\frac{2x\ln(x^2)}{1-x^2}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx-\int_{0}^{1}\frac{\ln x}{1-x}dx\\
&=-\frac{3}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx\\
&=-\frac{3}{4}\zeta(2)\\
&=-\frac{3}{4}\times -\frac{\pi^2}{6}\\
&=\frac{\pi^2}{8}\\
\end{align}
NB:
I assume that,
\begin{align} \zeta(2)=\frac{\pi^2}{6}\\
\end{align}
And,
\begin{align}\int_{0}^{1}\frac{\ln x}{1-x}dx&=\int_0^1\left(\sum_{n=0}^\infty x^n\right)\ln x\,dx\\
&=\sum_{n=0}^\infty \int_{0}^{1}x^n\ln x\,dx\\
&-=\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=-\zeta(2)
\end{align}
For $0\leq x<1$,
\begin{align}
\frac{1}{x^2-1}=\frac{x}{1-x^2}-\frac{1}{1-x}
\end{align}
</s>