My question is compute $$ \int\limits_{C(1, 1.5)} \frac{z^7+1}{z^2(z^4+1)} dz $$
where $C(1,1.5)$ is the circle $C$ centered at 1 with radius 1.5. So $C$ has $0$ and two of the four roots of $z^4+1$ in its interior. Let $r_1=e^{\pi i/4}$ and $r_2=e^{-\pi i/4}$ be the two roots in the interior. My calculations show that around $0$ the integral doesn't give any contribution (because derivative of $\frac{z^7+1}{z^4+1}$ at $0$ is $0$. At $r_1$ I'm calculating a contribution of $$2\pi i \frac{-r_1^3+1}{r_1^2\cdot (r_1^2+i)\cdot (r_1+r_1)} = \frac{-\pi i(r_2+1)}{2r_1}$$ since $r_1^2=i$, $-r_1^3=r_2$. Similarly I get the contribution around $r_2$ is $\frac{-\pi i(r_1+1)}{2r_2}$. Adding these I get $-\pi i/\sqrt{2}$, although the answer in the back of the book is $\pi i/\sqrt{2}$. Where am I going wrong, and is there an easier way? Is it possible someone could compute the integral in mathematica, maple, sage, etc? The answer in the back said it can be done by partial fraction decomposition.