\int_{-\infty}^{+\infty}|x|\frac{dx}{\pi(1+x^2)}=
 2\int_{0}^{+\infty}\frac{xdx}{\pi(1+x^2)}=(u=1+x^2)=
 \frac{1}{\pi}\int_{1}^{+\infty}\frac{du}{u}=