Find the value of the integral $\int_{c}\frac{3z^2+2}{(z-1)(z^2+9)}$ around the circle $c:|z-2| = 2$
In the answer I got $\operatorname{Res}_{z=1} \frac{z^3+2}{z^2+9} = \frac12$
therefore the integration is equal to $2\pi i*\frac1{2} = \pi i$
but how do we get $(z^3+2)/(z^2+9)$ ?
I remember that the residue theory is $\int_{c}f(z)dz = 2\pi i ∑\operatorname{Res}_{z_{k}}f(z)$
I think this problem has applied this theory, but I don't know how to get the Residue of $f(z)$
Because\begin{align}\operatorname{res}_{z=1}\frac{3z^2+2}{(z-1)(z^2+9)}&=\lim_{z\to1}(z-1)\frac{3z^2+2}{(z-1)(z^2+9)}\\&=\lim_{z\to1}\frac{3z^2+2}{z^2+9}\\&=\frac12.\end{align}