Question on calculating $\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$ My task is to calculate $$\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$$ using the Cauchy integral formula. My question is: Is there a simple trick or do I have to perform a partial fraction decomposition?

A partial fraction decomposition is not necessary. 
The function
\begin{align*}
f(z)=\frac{z^7 + 1}{z^2(z^4 + 1)}
\end{align*}
has a double pole at $z=0$ and simple poles at
$z\in\left\{e^{\frac{\pi i}{4}},e^{-\frac{\pi i}{4}},e^{\frac{3\pi i}{4}},e^{-\frac{3\pi i}{4}}\right\}$. The contour of integration $\partial B_{3/2}(1)$ is a circle with center $(1,0)$ and radius $\frac{3}{2}$. It encloses precisely the poles at
$0,e^{\frac{\pi i}{4}},e^{-\frac{\pi i}{4}}$.

We obtain
  \begin{align*}
\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz&=\int_{\partial B_{3/2}(1)}f(z)dz\\
&=2\pi i\left(\operatorname{res}\left(f,0\right)
+\operatorname{res}\left(f,e^{\frac{3\pi i}{4}}\right)
+\operatorname{res}\left(f,e^{-\frac{\pi i}{4}}\right)\right)\\
&=2\pi i\left(0+\frac{1}{4}\left(e^{\frac{3\pi i}{4}}+i\right)+\frac{1}{4}\left(-e^{\frac{\pi i}{4}}-i\right)\right)\\
&=2\pi i\left(-\frac{1}{2\sqrt{2}}\right)\\
&=-\frac{\pi i}{\sqrt{2}}
\end{align*}

