$$\int_C \frac{z+1}{z^2-2z} dz$$ 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}