<s>[INST] Show that $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ How do I show that: $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ using contours and residues
My attempt:
I know that the singular points are $i,-i,-1,1,0$
consider $f(z)= \frac{\ln z}{(z^2+1)(z^2-1)}$
and the branch $|z|>0$, $0<\theta<2\pi$ 
$u: z=r, \rho\le r \le R$ (u is the upper edge)
$-l: z=r, \rho\le r \le R$ (lower edge)
$\int_ufdz-\int_{-l}fdz=\int_\rho^R \frac{\ln r + i0}{(z^2+1)(z^2-1)}-\int_\rho^R \frac{\ln r + i2\pi}{(z^2+1)(z^2-1)}$
How do I continue from here? 
 [/INST] Observe that 
\begin{eqnarray*}
\frac{1}{(x^2+1)(x^2-1)}=\frac{1}{2(x^2-1)} + \frac{-1}{2(x^2+1)}
\end{eqnarray*}
So the original integral splits into the following two integrals
\begin{eqnarray*}
I_1=\int_{0}^{\infty} \frac{ln(x)}{2(x^2+1)} \\
I_2=\int_{0}^{\infty} \frac{ln(x)}{2(x^2-1)} .\\
\end{eqnarray*}
Both of these integrals can be evaluated by spliting the inteval into $[0,1]$ and $[1, \infty)$ and then doing the substitution $x=1/y$. The first integral is zero and the second integral will be 
\begin{eqnarray*}
I_1=0 \\
I_2=\int_{0}^{1} \frac{ln(x)}{(x^2-1)} dx \\
\end{eqnarray*}
Now geometrically expand $\frac{1}{(x^2-1)}$ and use
\begin{eqnarray*}
\int_{0}^{1} x^{n} \ln(x) dx =\frac{-1}{(n+1)^2} \\
I_2=-\int_{0}^{1} \sum_{n=0}^{\infty} x^{2n} ln(x) dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\
\end{eqnarray*}
It well known that $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\color{red}{\frac{\pi^2}{8}}$.
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