Originally Posted by

**kevinlightman** I am trying to integrate log(x)/( (1+x^2)^2) and I'm stuck at trying to find the residue of that function at i. I've tried expanding it to find the coefficient of (x-i) but I just get into a mess and as the pole is a double covert pole I don't think there is any nice formula that would lead me to the answer. Can anyone help please?

Let $\displaystyle \displaystyle f(z)=\frac{\log(z)}{(1+z^2)^2}$. Then, you noted that $\displaystyle f$ has pole of order $\displaystyle 2$ at $\displaystyle i$ so that

$\displaystyle \displaystyle \begin{aligned}\underset{z=i}{\text{Res }}f(z) &= \lim_{z\to i}\frac{d}{dz}\left((z-i)^2f(z)\right)\\ &= \lim_{z\to i}\frac{z-2z\log(z)+i}{z(z+i)^3}\\ &= \frac{\pi+2i}{8}\end{aligned}$