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.

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 
$$
\int_0^1\frac{\ln(x)}{x^2-1}dx=\frac{\pi^2}{24}+\frac{\pi^2}{12} = \frac{\pi^2}{8}
$$
