<s>[INST] 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.
 [/INST] We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$
We will generalize by introducing parameter $\alpha$ such that
$$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$
And we have $I(0)=0$
Then
$$I'(\alpha )=-\int_0^1\frac{1}{(1-\alpha x)(x-2)}\,\mathrm dx=\frac{1}{2\alpha -1}\left[\ln\left(\frac{x-2}{1-\alpha x}\right)\right]_0^1=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
And we have
$$I'(\alpha )=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+c$$
$$I(0)=\frac{1}{2}\text{Li}_2(-1)+c=0\implies c=-\frac{1}{2}\text{Li}_2(-1)=\frac{\pi^2}{24}$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+\frac{\pi^2}{24}$$
$$\begin{align}I(1)&=\frac{1}{2}\text{Li}_2(1)+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{12}+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{8}\\
\end{align}$$

$$I(1)=\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\frac{\pi^2}{8}$$

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