Let $C_a$ denote the circle of radius $a$ about the origin, compute
$$\int_{C_a}\frac{z^2+e^z}{z^2(z-2)}dz.$$
So I quickly noticed theres two cases for $a \in (0,2)$ and $a \in (2,\infty)$. So for $a \in (0,2)$, one has
\begin{align} \int_{C_a}\frac{z^2+e^z}{z^2(z-2)}dz&=2\pi i ((f'(\frac{z^2+e^z}{z-2}))\vert_{z=0})\\ &=2 \pi i (-\frac{3}{4})\\ &=-\frac{3\pi i}{2}. \end{align}
Since the winding number about $2$ is zero, the residue is zero as well.
But for $a>2$, its $2\pi i$ times the sum of the residue at $z=0,2$. For this, I get
\begin{align} \int_{C_a} \frac{z^2+e^z}{z^2(z-2)}dz&=2\pi i (-\frac{3}{4}+(\frac{z^2+e^z}{z^2})\vert_{z=2})\\ &=2\pi i (-\frac{3}{4}+\frac{4+e^2}{4})\\ &=\pi i (\frac{1+e^2}{2}). \end{align}
Is this correct?? Am I corrct in stating the answer is dependant on wether or not $a$ is greater than or less than $2$?? Just checking because the source I got this from said the answer for $a>2$ is $-\frac{\pi i}{2}$
I thought it might be instructive to present another approach. Note that we have
$$\begin{align} \frac{z^2+e^z}{z^2(z-2)}&=-\frac{z^2+e^z}{2z^2}-\frac{z^2+e^z}{4z}+\frac{z^2+e^z}{4(z-2)}\\\\ &=-\frac12-\frac{3e^z}{4z^2}-\frac{z}{4}+\frac{z^2+e^z}{4(z-2)} \end{align}$$
The only residue at $z=0$ is equal to $-3/4$ since $e^z=1+z+O(z^2)$. Hence, $2\pi i (-3/4)=-i3\pi/2$.
The residue at $z=2$ is $1+e^2/4$.
Therefore, we have
$$\oint_{|z|=a}\frac{z^2+e^z}{z^2(z-2)}\,dz=\begin{cases}-i\frac{3\pi}2&,a<2\\\\\ i\frac\pi2(1+e^2)&,a>2\end{cases}$$
The cases for $a=2$ must be interpreted in the sense of principal values. In that case, the principal value integral is simply the average of the results for $a<2$ and $a>2$.