<s>[INST] Complex integral involving Cauchy integral formula Where $C$  is the circle $|z|=\frac{3}{2}$, evaluate the following integral using the Cauchy integral formula. 
$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz$$
Clearly the simple poles at $z=\pm i$ are the ones that are inside the circle $C$, and the simple poles at $z=\pm 2$ are outside the circle.
I use partial fractions to simplify $$\frac{e^{z}}{z^{2}-4}$$ to get it equal to $$\frac{e^{z}}{4(z-2)}-\frac{e^{z}}{4(z+2)},$$ then define $$f(z)=\frac{e^{z}}{4},$$  which gives us $$\frac{f(z)}{z-2}-\frac{f(z)}{z+2},$$ then by the C.I.F I get that
$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz=\frac{\pi i}{2}[e^{i}-e^{-i}].$$ 
Could someone tell me if this is correct, thanks!
 [/INST] Write the integral as a sum of integrals around $\{i,-i\}$. These integrals are easily evaluated using the Cauchy integral formula:
$$
    \frac{1}{2\pi i}\oint_{|z-i|=1/2}\frac{1}{(z-i)}\frac{e^z}{(z+i)(z^2-4)}dz
  = \frac{e^i}{(i+i)(i^2-4)} \\
    \frac{1}{2\pi i}\oint_{|z+i|=1/2}\frac{1}{(z+i)}\frac{e^z}{(z-i)(z^2-4)}dz=\frac{e^{-i}}{(-i-i)((-i)^2-4)} \\
%   \frac{1}{2\pi i}\oint_{|z-2|=1/2}\frac{1}{(z-2)}\frac{e^z}{(z+2)(z^2+1)}dz=\frac{e^2}{(2+2)(2^2+1)} \\
%  \frac{1}{2\pi i}\oint_{|z+2|=1/2}\frac{1}{(z+2)}\frac{e^z}{(z-2)(z^2+1)}dz=\frac{e^{-2}}{(-2-2)((-2)^2+1)}
$$
The answer is the sum of these two integrals.
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