Integral imprópria

[latex]\\\mathrm{Integrando\ por\ partes:\int udv=uv-\int vdu}\\\\ \mathrm{u=arctan(x),dv=\frac{x}{(x^2+1)^2}\ \therefore\ du=\frac{1}{x^2+1},v=-\frac{1}{2(x^2+1)}}\\\\ \mathrm{\int_{0}^{\infty}\frac{xarctan(x)}{(x^2+1)^2}dx=-\frac{arctan(x)}{2(x^2+1)}+\frac{1}{2}\int \frac{1}{(x^2+1)^2}dx}\\\\ \mathrm{\int_{0}^{\infty}\frac{xarctan(x)}{(x^2+1)^2}dx=-\frac{arctan(x)}{2(x^2+1)}+\frac{1}{2}\left [\frac{x}{2(x^2 +1)}+\frac{1}{2}\overset{arctan(x)}{\overbrace{\mathrm{\int \frac{1}{x^2+1}dx}}} \right ]}\\\\ \mathrm{\int_{0}^{\infty}\frac{xarctan(x)}{(x^2+1)^2}dx=\left [\frac{x}{4(x^2+1)}+\frac{arctan(x)}{4}-\frac{arctan(x)}{2(x^2+1)} \right ]_{0}^{\infty}=\frac{\pi}{8}}[/latex]