Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$ 
Evaluate 
  $$\int_{|z|=3}  {1 \over P(z)}dz$$ 
  where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.

I proved that  $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are inside the ball.
$$P(z)=(z-9)(2z^4+2z^3+z^2-2z+2)=(z-9)(z-a_1)(z-\bar a_1)(z-a_2)(z-\bar a_2).$$ 
 A: By using the residue at infinity, we have that the sum of all residues of $1/P(z)$ at the poles is zero. 
Hence, since you already know that there the pole 9 is the only which is outside the the circle $|z|=3$, it follows that
$$\int_{|z|=3} {dz \over P(z)}=-2\pi i\cdot\mbox{Res}\left(\frac{1}{ P(z)},9\right)=-\frac{2\pi i}{P'(9)}=-\frac{2\pi i}{14645}.$$
P.S. By the way
$$P(z)=(z-9)(2z^4+2z^3+z^2-2z+2)=(z-9)(z^2+2z+2)(2z^2-2z+1)$$
so you can actually find the four poles inside $|z|=3$, but I think that the spirit of your exercise is to avoid the evaluation of four annoying residues.
A: The roots of $P(z)= 2z^4 + 2z^3+ z^2-2z+2 $ are $a_1 = -1-i$, $a_2 = \frac{-a_1}{2}$ , $\overline{a_1}$ and $\overline{a_2}$. Now by the residue Theorem, you get,
\begin{equation*}
\int_{|z|=3} \frac{dz}{P(z)} = 2i\pi \left( \frac{1}{(a_1 -9)P'(a_1)} + \frac{1}{(\overline{a_1}-9)P'(\overline{a_1})} + \frac{1}{(a_2-9)P'(a_2)} + \frac{1}{(\overline{a_2}-9)P'(\overline{a_2})} \right)
\end{equation*}
Since $P'$ has real coefficients, this sum is easier to compute....
A: Let
$$
P(z)=2z^5-16z^4-17z^3-11z^2+20z-18
$$
Since
$$
\begin{align}
&(-218840155-162129667x-67092610x^2+13150850x^3)P(z)\\
&-(3927717-68602966x-46831817x^2-17626794x^3+2630170x^4)P'(z)\\
&=3860568450
\end{align}
$$
$\gcd(P,P')=1$. Therefore, $P$ has no repeated roots.
$P(9)=0$ and $\frac{P(x)}{x-9}=2x^4+2x^3+x^2-2x+2$ implies that the remaining roots have absolute value no greater than $\frac{2\cdot\frac12+1\cdot\frac14+2\cdot\frac18+2\cdot\frac1{16}}{2}\cdot2=\frac{13}8$. Therefore, all the roots are simple and contained in $|z|\lt3$ except for the root at $z=9$. The sum of the residues of $\frac1{P(z)}$ is $0$ so the sum of the residues inside $|z|=3$ is the negative of the residue at $z=9$, which is
$$
-\frac1{P'(9)}=-\frac1{14645}
$$
Therefore,
$$
\int_{|z|=3}\frac{\mathrm{d}z}{2z^5-16z^4-17z^3-11z^2+20z-18}=-\frac{2\pi i}{14645}
$$
