$\oint_{C} \frac{2z^3+z^2+4}{z^4+4z^2}$
When using partial fractions I have $\frac{2z^3+z^2+4}{z^4+4z^2}$=$\frac{1}{z^2}+\frac{2z}{z^2+4}$
Then $\oint_{C} \frac{2z^3+z^2+4}{z^4+4z^2}dz=\oint_{C}\frac{1}{z^2}dz+\oint_{C}\frac{2z}{z^2+4}dz$
Now since $C$ is the circle of radius $4$ and center at $2$ we can see that the singular points are inside $C$.
The singular points are $z=0$, $z=2i$ and $z=-2i$
CAUCHY´S INTEGRAL FORMULA
Let $f(z)$ be analytic in a simply connected domain D. Then for any point $z_0$ in $D$ and any simple closed path $C$ in $D$ that encloses $z_0$
$\oint_{C}\frac{f(z)}{z-z_{0}}dz=2\pi i f(z_{0})$
DERIVATIVES OF AN ANALYTIC FUNCTION
If $f(z)$ is analytic in a domain $D$, then it has derivatives of all orders in D, which are then also analytic functions in $D$. The values of these derivatives at a point $z_{0}$ in $D$ are given by the formula
$\oint_{C}\frac{f(z)}{(z-z_{0})^{n+1}}dz=\frac{2\pi i }{n!}f^{n}(z_{0})$
So applying this to the integrals $\oint_{C}\frac{1}{z^2}dz$ and $\oint_{C}\frac{2z}{z^2+4}dz$ i get
$\oint_{C}\frac{1}{z^2}dz=-2\pi i f^{(1)}(0)=0$
and
$\oint_{C}\frac{2z}{z^2+4}dz=\oint_{C}\frac{2z}{(z-2i)(z+2i)}dz=-2\pi i [f(2i)+f(-2i)]$
where $f(2i)=\frac{2(2i)}{2i+2i}=1$ and $f(-2i)=\frac{2(-2i)}{-2i-2i}=1$
Thus
$\oint_{C}\frac{2z^3+2^2+4}{z^4+4z^2}dz=-2\pi i [1+1]=-4\pi i$.
But I have a huge doubt because apparently the result of the general is zero and I don't know how I can make the integral $\oint_{C}\frac{2z}{z^2+4}dz$ of zero.
Any suggestion? Or is there some mistake that I made to help me see