Determine $\displaystyle\int_{\gamma}{\frac{e^{z^{2}}}{z-1}dz}$, where $\gamma$ is the rectangule given by $x=0$, $x=3$, $y=-1$ e $y=1$.
My approach: If we consider the rectangule, then each side is given by $$\gamma_{1}:=z_{1}(t)=(3-i)t-(1-t)i$$
$$\gamma_{2}:=z_{2}(t)=(3+i)t-(1-t)(3-i)$$
$$\gamma_{3}:=z_{3}(t)=it+(3+i)(1-t)$$ $$\gamma_{4}:=z_{4}(t)=(1-t)i-it$$
My idea was use the Cauchy Integral Formula, because the function $f(z)=e^{z^2}$ is analytic over the rectangule, so
$$\displaystyle\int_{\gamma}{\frac{e^{z^{2}}}{z-1}dz}=2\pi i f(1)=2\pi ie$$
But I don't know how use the contour, thanks.
Your answer is correct. Dependence on the contour is only through the index. It is implicitly assumed that you go round the rectangle only once which makes the index of $1$ w.r.t. the contour equal to $1$.