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!
 A: Since the only singularities enclosed by $|z|=3/2$ are at $\pm i$, the function $f(z)=\frac{e^z}{z^2-4}$ is analytic in and on $|z|=3/2$.  Now, using partial fraction expansion, we have
$$\frac{1}{z^2+1}=\frac{1/i2}{z-i}-\frac{1/i2}{z+i}$$
Hence, from Cauchy's Integral Theorem we find 
$$\begin{align}
\oint_{|z|=3/2}\frac{e^z}{(z^2+1)(z^2-4)}\,dz&=\frac1{i2}\oint_{|z|=3/2}\frac{f(z)}{z-i}\,dz-\frac1{i2}\oint_{|z|=3/2}\frac{f(z)}{z+i}\,d\\\\
&z=\pi (f(i)-f(-i)\\\\
&=\pi \frac{e^{i}}{-5}-\pi \frac{e^{-i}}{-5}\\\\
&=\frac\pi5(e^{-i}-e^i)\\\\
&=-\frac{i2\pi\sin(1)}5
\end{align}$$
A: 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.
