Some help needed to evaluate the following integral using Residue theorem Some help needed to evaluate the following integral using Residue theorem.
The Integral of interest is as follow. 
$\int_C \frac{z^2}{(z^2+4)(z^2+9)}dz$
Where $C$ is the contour $|z| = 4$ traversed in the counterclockwise direction.
I wanted to use the following to evaluate, as it does help to simplify matters a bit. 
$\int_C f(z)dz = 2\pi i$Res$(\frac{1}{z^2}f(\frac{1}{z}), 0)$
So after the appropriate substitution I got the following
$\frac{1}{(1+4z^2)^2(\frac{1}{z^2} + 9)} = \frac{z^2}{(1+4z^2)^2(1+9z^2)}$  
But how do I find the residue (about $0$) for the "simplified" expression.
My attempts ended up in quite a mess, 
any help or insight is deeply appreciated.
 A: The answer is zero and it actually does not require to compute any residue. $f(z)=\frac{z^2}{(z^2+4)(z^2+9)}$ fulfills $\left|f(z)\right|\leq \frac{C}{|z|^2}$ as $|z|\to +\infty$, hence the sum of the residues of $f(z)$ is given by:
$$\frac{1}{2\pi i}\oint_{|z|=4}f(z)\,dz = \frac{1}{2\pi i}\oint_{|z|=M}f(z)\,dz $$
for any $M>3$, since the poles of $f(z)$ lie at $\pm 2i$ and $\pm 3i$. However,
$$\left|\oint_{|z|=M}f(z)\,dz\right|\leq\oint_{|z|=M}\frac{C}{|z|^2}\,dz = \frac{2\pi CM}{M^2} $$
is arbitrarily close to zero as $M\to +\infty$, hence the sum of the residues is zero.
A: Your idea of doing a "$w = 1/z$" substitution is very clever! As you hinted at, the integral we want is
$$ \int_{|w| = \frac 1 4} \frac{w^2}{(1+4w^2)(1+9w^2)}\frac {dw}{w^2}.$$
(Note that $dz = -dw / w^2$, hence the extra factor of $w^2$ in the denominator, which matches your formula. If we drop the minus sign in $-dw / w^2$, then the contour $|w| = \frac 1 4$ is to be traversed in the anticlockwise direction.)
So we need the residues of
$$ \frac{w^2}{(1+4w^2)(1+9w^2)} \times \frac {1}{w^2} = \frac{1}{(1+4w^2)(1+9w^2)}$$
at all of its poles inside the region $|w| < \frac 1 4$.
Your question about the "residue at $w = 0$" does not apply, because the integrand does not have a pole at $w = 0$.
It does have poles at $w = \pm i / 2$ and $w = \pm i / 3$, but since these poles lie outside the region $|w| < \frac 1 4$, they don't count either.
The conclusion is that the integrand is holomorphic in the region $|w| < \frac 1 4$: it has no poles in this region at all! Therefore, the integral is zero.
A: Let
$f(z) := z^2 /(z^2+4)(z^2+9)$ = $ \frac{z^2}{({z+2i})({z-2i})({z+3i})({z-3i})}$,
so f has simple poles at -2i, 2i, 3i and -3i.
Now, our contour, a circle centred at 0 of radius 2, contains none of these poles, so the sum of the residues inside this contour is 0.
Then, by the residue theorem, we have your integral equals 2pi*0 = 0
