Contour integral of $\int_\gamma \frac{dz}{(2z+1)(z+3)^2}$

Question

Im a little confused by the following integral question

Let $\gamma$ be the unit circle in $\mathbb{C}$ traversed in the anti-clockwise direction.

$\displaystyle\int_\gamma \dfrac{1}{(2z+1)(z+3)^2}$

z=-3 is outside the circle... and z=-$\frac{1}{2}$ is inside.

Hence.. $\displaystyle\int_\gamma \dfrac{1}{(2z+1)(z+3)^2}$ = $\displaystyle\int_\gamma \dfrac{\dfrac{1}{(z+3)^2}}{(2z+1)}$

And hence $f(x)=\dfrac{1}{(x+3)^2}$ and $f_o=\frac{1}{z}$? This is where im getting a little confused....

Answer 1

Hint! Cauchy Residue theorem says that your integral is $2\pi i \sum R_i$ where $R_i$ are the residues inside the closed contour, in you case the only pole inside is $-1/2$, so just take the residue at this point and multiply with $2 \pi i$