Evluate $\int_{|z|=3}\frac{1}{z^5(z^2+z+1)}\ dz$

The poles of the function  $f(z)=\frac{1}{z^5(z^2+z+1)}$ are

\[-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i,  \text{and}   0.\]. 
Since all poles are inside $|z|=2$, by the residue theorem we have

$$\tag 1 \int_{\{|z|=3\}} \frac{1}{z^5(z^2+z+1)}\, dz=\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz$$ 
for any $R>2.$

But
\[|\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz| \leq 2\pi R \frac{1}{R^5(R^2+R+1)} \rightarrow 0,\]
as $R \rightarrow \infty$. Thus 

\[ \int_{\{|z|=3\}} \frac{1}{z^5(z^2+z+1)}\, dz=\lim_{R \rightarrow \infty}\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz=0.\]