I'm a little stuck with this one. I've found the singularities to be at $\pm 2i$ and $0$ (branch point). So far, using a branch cut at $2\pi$ I've found that
$$\int\_{0}^{\infty}\frac{\ln(2r)}{4+r^2}dr+\int\_{0}^{\infty}\frac{\ln(2r)+2\pi i}{4+r^2}dr = 2I+\int\_{0}^{\infty}\frac{2\pi i}{4+r^2}dr$$
And 
$$\int\_{0}^{\infty}\frac{2\pi i}{4+r^2}dr = \text{Res}(r=2i) \\ = -2\pi^2\lim\_{r\to2i}\frac{r-2i}{(r-2i)(r+2i)} = \frac{i}{2}\pi^2$$
My problem is that the answer I found is completely imaginary, and I'm not sure how that's possible given that the original function is real. Any help is appreciated.