$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$
\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}
The substitution $x\mapsto\coth(x)$ works out quite well. From hereon we get
$$-\int_1^\infty\frac{\log(x)}{1-x^2}\mathrm dx=\int_\infty^0\log(\coth(x))\mathrm dx=\int_0^\infty\log(\coth(x))\mathrm dx$$
Since the hyperbolic cotangent is defined in terms of exponentials we may further rewrite this whole as
\begin{align*} &\int_0^\infty\log(\coth(x))\mathrm dx=\int_0^\infty\log\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)\mathrm dx = \int_0^\infty\log\left(\frac{1+e^{-2x}}{1-e^{-2x}}\right)\mathrm dx\\ =&\int_0^\infty\log\left(1+e^{-2x}\right)-\log\left(1-e^{-2x}\right)\mathrm dx=\int_0^\infty\sum_{n=0}^\infty(-1)^{n+1}\frac{e^{-2nx}}{n}+\sum_{n=0}^\infty\frac{e^{-2nx}}{n}\mathrm dx \end{align*}
Due the monotone dominate convergence theorem we are allowed to switch the order of integration and summation and integration which further gives us
\begin{align*} &\int_0^\infty\sum_{n=0}^\infty(-1)^{n+1}\frac{e^{-2nx}}{n}+\sum_{n=0}^\infty\frac{e^{-2nx}}{n}\mathrm dx=\sum_{n=0}^\infty(-1)^{n+1}\int_0^\infty\frac{e^{-2nx}}{n}\mathrm dx+\sum_{n=0}^\infty\int_0^\infty\frac{e^{-2nx}}{n}\mathrm dx\\ =&\sum_{n=0}^\infty(-1)^{n+1}\left[-\frac{e^{-2nx}}{2n}\right]_0^\infty+\sum_{n=0}^\infty\left[-\frac{e^{-2nx}}{2n}\right]_0^\infty=\sum_{n=0}^\infty\frac{(-1)^{n+1}}{2n^2}+\sum_{n=0}^\infty\frac1{2n^2}=\frac12[\eta(2)+\zeta(2)]\\ =&\frac12\left[\frac{\pi^2}{12}+\frac{\pi^2}6\right]=\frac12\left[\frac{\pi^2}4\right]=\frac{\pi^2}8 \end{align*}
$$\therefore~\int_1^\infty\frac{\log(x)}{x^2-1}\mathrm dx~=~\int_0^\infty\log(\coth(x))\mathrm dx~=~\frac{\pi^2}8$$
$\zeta(s)$ and $\eta(s)$ are the Riemman Zeta Function and the Dirichlet Eta Function respectively. Their series definitions and the relation $\eta(s)=(1-2^{1-s})\zeta(s)$ were used aswell as the well-known value of $\zeta(2)$. Please ask if something is unclear. Overall it is just another way of gaining the series representation of your integral.
Basically it is the same question as here: Integral of $\ln(\tanh(x))$ since you get your value by multiplying the expression by minus one due the relation between the hyperbolic tangent and cotagent function.
I will provide two different methods. The first relies on properties of the polygamma function, the second converts the integral to a double integral first before evaluating it.
Let $$I = \int_1^\infty \frac{\ln x}{x^2 - 1} \, dx.$$
Method 1 - A polygamma approach
By enforcing a substitution of $x \mapsto 1/x$ we see that $$I = \int_0^1 \frac{\ln x}{x^2 - 1} \, dx.$$
From the following integral representation for the digamma function $\psi (x)$, namely $$\psi (x + 1) = -\gamma + \int_0^1 \frac{1 - t^x}{1 - t} \, dt,$$ where $\gamma$ is the Euler–Mascheroni constant, on differentiating with respect to $x$ we have $$\psi^{(1)} (x + 1) = - \int_0^1 \frac{t^x \ln t}{1 - t} \, dt.$$ Here $\psi^{(1)} (z)$ denotes the trigamma function. Substituting $t = u^2$, $dt = 2u \, du$ on finds $$\psi^{(1)} (x + 1) = 4 \int_0^1 \frac{u^{2x + 1} \ln u}{u^2 - 1} \, du.$$ Setting $x = -1/2$ then yields $$\int_0^1 \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \psi^{(1)} \left (\frac{1}{2} \right ).$$ Here the dummy variable $u$ has been reverted back to $x$.
To find the value for the trigamma function at $x = 1/2$ we note that for the polygamma function one has (see Eq. (16) here) $$\psi^{(n)} \left (\frac{1}{2} \right ) = (-1)^{n + 1} n! (2^{n + 1} - 1) \zeta (n + 1).$$ Here $\zeta (z)$ denotes the Riemann zeta function. Setting $n = 1$ yields $$\psi^{(1)} \left (\frac{1}{2} \right ) = 3 \cdot \zeta (2) = 3 \cdot \frac{\pi^2}{6} = \frac{\pi^2}{2}.$$ Thus $$\int_1^\infty \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \cdot \frac{\pi^2}{2} = \frac{\pi^2}{8}.$$
Method 2 - Using a double integral
The problem the first method suffers from is its heavy reliance on a knowledge of the polygamma function. In this second approach, knowing any properties for the polygamma and zeta functions are completely avoided altogether.
Note that as $$\int_0^\infty \frac{\ln x}{x^2 - 1} \, dx = \int_0^1 \frac{\ln x}{x^2 - 1} \, dx + \int_1^\infty \frac{\ln x}{x^2 - 1} \, dx,$$ we have $$I = \frac{1}{2} \int_0^\infty \frac{\ln x}{x^2 - 1} \, dx = \frac{1}{4} \int_0^\infty \frac{\ln (x^2)}{x^2 - 1} \, dx. \tag1$$ Observing that $$\ln (x^2) = \int_0^\infty \left (\frac{x^2}{1 + x^2 t} - \frac{1}{1 + t} \right ) \, dt,$$ we can rewrite (1) as \begin{align} I &= \frac{1}{4} \int_0^\infty \int_0^\infty \left (\frac{x^2}{1 + x^2 t} - \frac{1}{1 + t} \right ) \frac{1}{x^2 - 1} \, dt \, dx\\ &= \frac{1}{4} \int_0^\infty \frac{1}{1 + t} \int_0^\infty \frac{1}{1 + x^2 t} \, dx \, dt, \end{align} after the order of integration has been changed. Evaluating we have \begin{align} I &= \frac{1}{4} \int_0^\infty \frac{1}{\sqrt{t} (1 + t)} \big{[} \tan^{-1} (x \sqrt{t}) \big{]}_0^\infty \, dt\\ &= \frac{\pi}{8} \int_0^\infty \frac{dt}{\sqrt{t} (1 + t)}\\ &= \frac{\pi}{4} \int_0^\infty \frac{dy}{1 + y^2} \qquad \text{(let $t = y^2$)}\\ &= \frac{\pi}{4} \big{[} \tan^{-1} y \big{]}_0^\infty\\ &= \frac{\pi}{4} \cdot \frac{\pi}{2}\\ &= \frac{\pi^2}{8}, \end{align} as expected.