complex analysis - Showing $int_{0}^{infty} frac{1}{(x^2+1)^2(x^2+4)}=frac{pi}{18}$ via contour integration

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

Answer

Consider the contour $C$ that spans along $-R$ to $R$ and around the arc $Re^{i\theta}$ for $0\le\theta\le \pi$.

Letting

$$f(z):=\frac{1}{(z^2+1)^2(z^2+4)}=\frac{1}{(z+i)^2(z-i)^2(z+2i)(z-2i)}$$

and we see the poles are located at $\pm i$ and $\pm 2i$. Letting $R \to \infty$, it is very clear that the denominator explodes, causing the integral around the arc to disappear. Then

$$\oint_C f(z)\, dz = 2\pi i(\operatorname*{Res}_{z = i}f(z) + \operatorname*{Res}_{z = 2i}f(z))$$

because $2i$ and $i$ are the only poles in $C$.
The pole of $i$ is of order 2:

$$
\operatorname*{Res}_{z = i}f(z) =
\lim_{z \to i} \frac{1}{1!}\frac{d}{dz} (z-i)^2 f(z)=
\lim_{z \to i} \frac{d}{dz}\frac{1}{(z+i)^2(z^2+4)}=
\lim_{z \to i} \frac{2(2z^2 +iz+4)}{(i+z)^3(4+z^2)^2}=-\frac{i}{36}
$$

The pole of $2i$ is simple:

$$
\operatorname*{Res}_{z = 2i}f(z) =
\lim_{z \to 2i} (z-2i)f(z) = \frac{1}{(-4+1)^2(2i+2i)}=-\frac{i}{36}
$$

So finally

$$
\int_0^\infty f(x)\, dx = \frac{1}{2}\int_{-\infty}^\infty f(x)\, dx = \pi i\left(-\frac{i}{36}-\frac{i}{36}\right) = \frac{\pi}{18}
$$