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**Dinkydoe** I need to show that $\displaystyle \int_{0}^{\infty}\frac{\log(x)}{(1+x^2)^2}=-\frac{\pi}{4}$

If we complexify the integrand and make a sort of pacman-contour $\displaystyle \mu$

$\displaystyle \mu_1 = t+ ic_0$ with $\displaystyle 0\leq t\leq R$

$\displaystyle \mu_2 = Re^{i\pi t}$ with $\displaystyle t_0\leq t \leq 2\pi-t_0$

$\displaystyle \mu_3 = R-t-ic_0 $ with $\displaystyle 0\leq t\leq R $

$\displaystyle \mu_4 = re^{i\pi t}$ with $\displaystyle 2\pi- t_0\leq t \leq t_0 $

As $\displaystyle R\to \infty, r\to 0$ we let $\displaystyle t_0,c_0\to 0$

This contour contains the 2 second order singularities $\displaystyle p= \pm i$