I was asked to verify Cauchy's integral theorem when $f(z)=\frac{1}{z^2+2}$ and $\gamma$ is the circle $|z+1|=\frac{1}{2}$. I have tried various ways and eventually it leads to complicated integration.
This is my approach, but I don't know how to continue on. We parametrize the circle as $\gamma(t)=-1+\frac{1}{2}e^{it}$. Then $\gamma^{'}(t)=\frac{1}{2}ie^{it}$. Then $$\oint_{\gamma}f(z)dz=\int_{0}^{2\pi}\frac{1}{(-1+\frac{1}{2}e^{it})^2+2}\frac{1}{2}ie^{it}dt=\int_0^{2\pi}\frac{2ie^{it}}{e^{2it}-4e^{it}+12}dt$$ and then I was stucked at here.
Is there any hint so that I can proceed this question ? Thank you.