I want to show that: $\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$ so considering: $\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}$ where gamma is the curve going from $0$ to $-R$ along the real axis, from $-R$ to R via a semi-circle in the upper plane and then from $R$ to 0 along the real axis.
Using the residue theorem we have that: $\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}=2\pi i \sum Res$ so re-writing the integrand as $\displaystyle\frac{1}{(z-2i)(z+2i)(z+i)^2(z-i)^2}$
we can see that there is two simple poles at $2i$,$-2i$ and two poles of order 2 at $i$,$-i$. Calculating the residues: $Res_{z=2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z+2i)(z+i)^2(z-i)^2}=\frac{1}{36i}$
$Res_{z=-2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z-2i)(z+i)^2(z-i)^2}=\frac{-1}{36i}$
$Res_{z=i}\lim_{z\rightarrow i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z+i)^2}=\frac{2i}{36}+\frac{2}{24i}$
$Res_{z=-i}\lim_{z\rightarrow -i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z-i)^2}=\frac{-2i}{36}+\frac{-2}{24i}$
But now the sum of the residues is 0 and so when I integrate over my curve letting R go to $\infty$ (and the integral over top semi-circle goes to 0) I will just get 0?
Not sure what I've done wrong? Thanks very much for any help