a question about Cauchy integral formula

Question

I'm new in the complex analysis and I'm stuck with this integral :

$I=\displaystyle \int_{|z|=4} \frac{\mathrm{d}z}{(z^2+9)(z+9)} $

the exercise is about Cauchy integral, I don't want the whole solution, just give me a hint (Please don't post fully worked solutions)

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what I have done : using partial fraction we get :

$I=\displaystyle \frac{1}{90} \int_{|z|=4} \frac{1}{9+z} + \frac{(9-z)}{9+z^2} \mathrm{d}z$

I'm trying to do this : $\displaystyle \int_{|z|=4} \frac{\mathrm{d}z}{9+z} $.

but $|9|>4$, how to do the integration in this case ?

Answer 1

Note that $(z^2 + 9)(z+9) = (z-3i)(z+3i)(z+9)$, so the integrand has simple poles at $3i, -3i,$ and $-9$. Figure out which of these are in the region $|z| \leq 4$ and then apply Cauchy's Integral Formula.

Answer 2

The function $1/(9+z)$ is a holomorph function on the integration domain $\{|z| = 4\}$, hence by Cauchy integral formula its inegral vanishes: $$\int_{|z|=4} \frac{dz}{9 + z} = 0 $$

How would you proceed with the second term?