$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
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?

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.
