$\int z{P'(z) \over P(z)}dz$ in {$z$ complex / $ |z|=3 $} where $P(z)=z⁵+z⁴-9z²+7$

Question

$\int z{P'(z) \over P(z)}dz$

I proved that $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, so all the poles of $ z{P'(z) \over P(z)}$ are inside the ball. What should I do next? How do I find the residues?

Edit: The zeroes of $P(z)$ are poles of $z{P'(z) \over P(z)}$ of order 1. If $a_1$ is a zeroe of P then $res({P'(z) \over P(z)},a_1)=m_1$ where $m_1$ is the multiplicty of the zeroe in $a_1$. Then the integral equals $2 \pi i \sum a_i m_i$.

Edit 2: $\sum a_i m_i$ is minus the coefficient of order 4 of $P(z)$, which is 1, then $\int z{P'(z) \over P(z)}dz = -2 \pi i$

Answer 1

$P(z)\neq 0$ for all $z$ outside of the ball of radius 3, so all the poles of $ z{P'(z) \over P(z)}$ are inside the ball. The zeroes of $P(z)$ are poles of $z{P'(z) \over P(z)}$ of order 1. If $a_1$ is a zeroe of P then $res({P'(z) \over P(z)},a_1)=m_1$ where $m_1$ is the multiplicty of the zeroe in $a_1$. Then the integral equals $2 \pi i \sum a_i m_i$.

$\sum a_i m_i$ is minus the coefficient of order 4 of $P(z)$, which is 1, then $\int z{P'(z) \over P(z)}dz =- 2 \pi i$