From Cauchy's Integral Formula, we can define $f\left(z\right) = z^{2} + z$, and find for any contour $C$ on the right part of the semi-plane $\Re\left(z\right) > 0$ that \begin{align*} \int_{C}\left(\frac{1}{z} + \frac{1}{z^{2}}\right)\mathrm{d}z & = \int_{C}\frac{z^{2} + z}{z^{3}}\,\mathrm{d}z = \int_{C}\frac{f\left(z\right)} {\left(z - 0\right)^{3}}\,\mathrm{d}z = \pi\mathrm{i}f''\left(0\right) = 2\pi\mathrm{i} \end{align*}
This should work since $z\not= 0$ because $\Re\left(z\right)>0$. However, if I want to calculate
\begin{align*} \int_{1 - \mathrm{i}}^{1 + \,\sqrt{\,{3}\,}\,\mathrm{i}} \left(\frac{1}{z} + \frac{1}{z^{2}}\right)\mathrm{d}z \end{align*}
Would the two integrals have the same value if the path intersect both points $?$. Or in the definite integral would be necessary to just integrate as the usual $?$.
In Cauchy's Integral Formula, the contour $C$ is the boundary of a disk in the complex plane that contains the singularity of the integrand. So, in the example here, we have
$$\oint_C \left(\frac1z+\frac1{z^2}\right)\,dz=2\pi i$$
if $z=0$ is interior to the disk bounded by $C$.
If $C$ is not the boundary of any disk for which $z=0$ is an interior point, then Cauchy's Integral Theorem guarantees that the integral over $C$ is $0$.