Contour integral of analytic function with singularity

I am supposed to integrate $f(z)=$$\frac{5}{z}$ from -3 t0 3 but I am having trouble understanding how to do this. I’ve done the integration the “hard” way by using parametrizations but now I need to use the fact that $f(z)$ is analytic everywhere except at $z=0$ and a different method to perform the integration but I am stumped.

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There is not quite enough information to answer this question, although you can narrow it down a bit. For example, all paths that stay in the upper half plane will give the same answer, which will be the same as if you go along the upper semicircle: $C: z(t) = 3e^{it}, \pi \geq t \geq 0$
$$\int_C \frac{5}{z}dz = \int_\pi^{0} 5\frac{3ie^{it}}{3e^{it}}dt = -5\pi i $$

If the path is completely in the lower half plane, the answer will be the same as integrating along the lower semicircle. $C: z(t) = 3e^{it}, \pi \leq t \leq 2\pi$

$$\int_C \frac{5}{z}dz = \int_\pi^{2\pi} 5idt = 5\pi i $$

There are some other weird paths that give any value of $5\pi i + 10n\pi i$. For example, if circle the origin clockwise $n$ times and then hit 3, the value will be $5\pi i – 10n\pi i$