Q. Consider the function $$F(z)=\int_{1}^{2} \frac {1}{(x-z)^2}dx, {\text {Im}(z) \gt 0}$$ Then there is a meromorphic function function $G(z)$ on $\Bbb C$ that agrees with $F(z)$ when ${\text {Im}(z) \gt 0}$, such that
A) $1,\infty$ are poles of $G(z)$.
B) $0,1,\infty$ are poles of $G(z)$.
C) $1,2$ are poles of $G(z)$.
D) $1,2$ are simple poles of $G(z)$.
I solved the definite integral and got $F(z)=\frac {1}{(z-1)(z-2)}$. So options C & D are correct. But I am not able to think about $0$ and $\infty$ being the poles. In what direction should I think to solve this?
Thank you.