The Monster Integral Strikes Again!

Split the integral as follows:

$\displaystyle \int_0^1 \frac{x(1-x^2)}{(1+x^2)^2}\ln\left(\frac{1+x}{1-x}\right)\,dx+\int_1^{\infty} \frac{x(1-x^2)}{(1+x^2)^2}\ln\left(\frac{x+1}{x-1}\right)\,dx$

In the second integral, perform the transformation $x \mapsto \dfrac{1}{x}$ to obtain:

$\displaystyle \int_0^1 \frac{1}{x}\left(\frac{1-x^2}{1+x^2}\right)^2\ln\left(\frac{1-x}{1+x}\right)\,dx$

Next, use the substitution $\dfrac{1-x}{1+x}=t \Rightarrow dx=\dfrac{-2}{(1+t)^2}\,dt$ i.e

$\int_0^1 \frac{1+t}{1-t}\frac{4t^2}{(1+t^2)^2}\ln t \frac{2}{(1+t)^2}\,dt$ $=8\int_0^1 \frac{t^2}{(1-t^2)(1+t^2)^2}\ln t\,dt=I$

From integration by parts and using the following result:

$\displaystyle \int \frac{t^2}{(1-t^2)(1+t^2)^2}\,dt=\frac{1}{8}\left(\frac{-2t}{1+t^2}-\ln(1-t)+\ln(1+t)\right)+C$

$\displaystyle \Rightarrow I=\left(\left(\frac{-2t}{1+t^2}-\ln(1-t)+\ln(1+t)\right)\ln t\right|_0^1-\int_0^1 \left(\frac{-2t}{1+t^2}-\ln(1-t)+\ln(1+t)\right)\frac{1}{t}\,dt$

Notice that the first term is zero i.e

$\displaystyle I=\int_0^1 \frac{2}{1+t^2}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt-\int_0^1 \frac{\ln(1+t)}{t}\,dt$

$\Rightarrow I=J_1+J_2-J_3$

$\displaystyle J_1=\int_0^1 \frac{2}{1+t^2}=2\left(\arctan t\right|_0^1=\frac{\pi}{2}$

$\displaystyle J_2=\int_0^1 \frac{\ln(1-t)}{t}\,dt=-\sum_{k=1}^{\infty} \frac{1}{k}\int_0^1 t^{k-1}\,dt=-\sum_{k=1}^{\infty} \frac{1}{k^2}=-\frac{\pi^2}{6}$

$\displaystyle J_3=\int_0^1 \frac{\ln(1+t)}{t}\,dt=-\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\int_0^1 t^{k-1}\,dt=-\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}=\frac{\pi^2}{12}$

Hence,

$\displaystyle I=\frac{\pi}{2}-\frac{\pi^2}{6}-\frac{\pi^2}{12}=\boxed{\dfrac{\pi}{2}-\dfrac{\pi^2}{4}}$