Computing $\int _C \frac {1}{z^3(z-1)^2}$, $C: |z-2|=5$ How do I compute $\int _C \frac {1}{z^3(z-1)^2}$, $C: |z-2|=5$?
I can't seem to use Cauchy's Formula, because both $0$ and $1$ are in the formula. 
There is this theorem, saying that $\int _C= \int_{C_1}+\int_{C_2}$, $C_1,C_2$ being the circles around the points where the function is not holomorphic, but I'm not sure how to use it here. It seem not intuitive that the function itself remains the same when used on $C_1,C_2$.
Anyone showing me how to deal with these problems will be extremely helpful, as this entire subject quite unclear to me.
Thank you for your time!  
Edit: The theorem I referred to above, as formulated in class (I translated it to English):

Let $U \subseteq \mathbb C$ be an open set. $\gamma$ is a closed simple curve in $U$ so that the interior of $\gamma$ ($int(\gamma)$) is entirely in $U$. Let $f$ be holomorphic in $U$ except a finitely many points $z_1, \cdots, z_n \in int(\gamma)$. Let there be circles $C_1, \cdots, C_n$, so that circle $C_i$ is centered at $z_i$, and for every $i \neq j$, $C_i \cap C_j= \emptyset$ and $int (C_i) \cap int(C_j)=\emptyset$, then
  $\int _\gamma f(z)dz=\Sigma_{i=1}^n \int_{C_i}f(z)dz$   

 A: Let $f=\frac{1}{z^3(z-1)^2}$. 
Then by the Residue Theorem, for $C: |z-2|=5$,
$\int _C \frac {1}{z^3(z-1)^2}=2\pi i(\text{Res}(f,0)+\text{Res}(f,1))$.
A quick calculation of the residues should give $\text{Res}(f,1)=-3$ and $\text{Res}(f,0)=3$.
So $\int _C \frac {1}{z^3(z-1)^2}=2\pi i(\text{Res}(f,0)+\text{Res}(f,1))=0$
So it looks like if we use your theorem we want to find $r_1,r_2>0$ so that 
$|z-1|=r_1$ and $|z|=r_2$ don't intersect as circles, but also that one does not lie in the other. Let's call the circles $C_1$ and $C_2$ respectively.
Let's choose $r_1=r_2=\frac{1}{4}.$
Then $C_1$ and $C_2$ will satisfy the properties of the theorem. 
\begin{align*}\int _C \frac{1}{z^3(z-1)^2}dz&=\int_{C_1}\frac{1}{z^3(z-1)^2}dz+\int_{C_2}\frac{1}{z^3(z-1)^2}dz
\end{align*}
We can rewrite the right-hand-side and use Cauchy's formula to solve the last line.
\begin{align*}
\int _C \frac{1}{z^3(z-1)^2}dz&=\int_{C_1}\frac{1}{z^3(z-1)^2}dz+\int_{C_2}\frac{1}{z^3(z-1)^2}dz\\
&=\int_{|z-1|=r_1}\frac{\frac{1}{z^3}}{(z-1)^2}dz+\int_{|z|=r_2}\frac{\frac{1}{(z-1)^2}}{z^3}dz\\
\end{align*}
