I am trying to evaluate $\frac{1}{2\pi i}\oint_C\frac{e^{et}}{(z^2+1)^2}dz$, where $C$ is $|z|=3$, knowing that it should be $\frac{1}{2}(\sin t-t \cos t)$. Here is what I've done:
$$f^{(n)}(a)=\frac{n!}{2 \pi i}\oint_C\frac{f(z)}{(z-a)^{n+1}}dz$$
$$\frac{1}{2\pi i}\oint_C\frac{e^{et}}{(z^2+1)^2}dz=\frac{1}{2\pi i}\oint_C\frac{\frac{e^{et}}{(z-i)^2}}{(z+i)^2}dz+\frac{1}{2\pi i}\oint_C\frac{\frac{e^{et}}{(z+i)^2}}{(z-i)^2}dz$$
$n=1$, $a_1=-i$, $a_2=i$
$f_1(z)=\frac{e^{zt}}{(z-i)^2}$, $f_2(z)=\frac{e^{zt}}{(z-i)^2}$
$f'_1(z)=\frac{te^{zt}(z-i)^2-e^{zt}2(z-i)}{((z-i)^2)^2}$, $f'_2(z)=\frac{te^{zt}(z+i)^2-e^{zt}2(z+i)}{((z+i)^2)^2}$
$f'_1(-i)=\frac{e^{-it}(-t+i)}{4}$, $f'_2(i)=-\frac{e^{it}(t+i)}{4}$
$$\frac{e^{-it}(-t+i)}{4}-\frac{e^{it}(t+i)}{4}$$
$$\frac{e^{-it}(-t+i)}{4}-\frac{e^{it}(t+i)}{4}=\frac{1}{4} \left(-i(e^{it}-e^{-it})-t(e^{it}+e^{-it}) \right)=$$
$$=\frac{1}{4} \left(\frac{1}{i}(e^{it}-e^{-it})-t(e^{it}+e^{-it}) \right)=\frac{1}{2} \left(\frac{e^{it}-e^{-it}}{2i}-t\frac{e^{it}+e^{-it}}{2} \right)=\frac{1}{2} (\sin t-t \cos t)$$