Are these the correct residues? $$\int_C \frac{z+1}{z^2-2z} dz$$ for the circle of $\lvert z \rvert = 3 $. Poles are obviously  at $ z = {0,2}$. Can I calculate the residues by viewing the fraction in the integral as either $$\int_C \frac{\frac{z+1}{z}}{z-2} dz $$$$ \int_C \frac{\frac{z+1}{z-2}}{z} dz$$ and plug into 2 and 0 into those numerators respectively? That would yield a final answer of $2\pi i * (\frac{3}{2} + \frac{1}{-2}) = \pi i$.
Does this look right? I'm new to residues and want to make sure I'm on the right track.

You have an addition mistake in your solution.
\begin{align}
\oint\frac{z+1}{z(z-2)}dz &= \oint\frac{(z+1)/z}{z-2}dz+\oint\frac{(z+1)/(z-2)}{z}dz\\
&= 2\pi i\bigl[f_1(2) + f_2(0)\bigr]
\end{align}
where $f_1(z) = \frac{z-1}{z}$ and $f_2(z)=\frac{z+1}{z-2}$. Then 
$$
\oint\frac{z+1}{z(z-2)}dz = 2\pi i(3/2-1/2) = 2\pi i
$$
We could also do this problem using Residue theory. We have simple poles inside the contour at $z=0,2$.
Then
\begin{align}
\oint\frac{z+1}{z(z-2)}dz &= 2\pi i\sum\text{Res}\\
&=2\pi i\biggl[\lim_{z\to 0}z\frac{(z+1)}{z(z-2)}+\lim_{z\to 2}(z-2)\frac{(z+1)}{z(z-2)}\biggr]\\
&= 2\pi i(-1/2+3/2)\\
&= 2\pi i
\end{align}
