Let \gamma be the circle |z-(1+i}|=2 and \int_{\gamma}\frac{1}{(z^2-4)(z^{2}+4)}dz=\frac{1}{32}\int_{\gamma}\frac{1}{z-2}-\frac{1}{8}\int_{\gamma}\frac{1}{z+2}dz-\frac{1}{32}\int_{\gamma}\frac{1}{z^{2}+4}dz (after PFD).

First thing I did was let

\gamma(t)=2e^{it}-\sqrt{2}e^{i\frac{\pi}{4}} for t\in{[0,2\pi]}. Then, \gamma\\'(t)=2ie^{it} for t\in{[0,2\pi]}.

Although, I am not sure I need this.

Also, the funtion g(z)=\frac{1}{(z^2-4)(z^{2}+4)} is undefined at the point \{-2i,-2,2,2i\} which are contained in \gamma.

Here is where my thoughts start breaking down. As I understand it, I can use Cauchy's integral formula to integrate around these bad areas.

So, for the first integral (looking at RHS) I see I have a bad point at z=2. So, I can say

\int_{\gamma}\frac{1}{z-2}dz=\int_{\gamma}\frac{f(z)}{z-2}dz where f(z)=\frac{1}{(z+2)(z^2+4)}. Thus,

\int_{\gamma}\frac{1}{z-2}dz=2\pi\\if(2)=\pi\\i

I would then do the same for the other chunks?

Note: The way I understand Cauchy's integral forumal is, if we do have a "bad" point in our integrand, we can still compute find the integral by creating smaller and smaller circles around that point. So, the thereom tells use we will still get the same integral as if we were integrating along \gamma.

Something like that....

On the other hand, if each integrand was analytic on whatever part of the plane, then everything would go to zero.

Any help would be great.