Let $\Gamma$ be the circle of radius 1,3 and 5 respectively. Evaluate
$$\int_\Gamma \frac{1}{(z+4)^3(z-2i)}dz$$
Attempt
$$\int_\Gamma \frac{1}{(z+4)^3(z-2i)}dz=\int_\Gamma -\frac{4+2i}{(z+4)^3}+\frac{1}{(z-2i)}dz$$
For $\Gamma$ a circle of radius 1 we have no isolated singularites within the loop and so, by Cauchy's theorem, we have
$$\int_\Gamma \frac{1}{(z+4)^3(z-2i)}dz=0$$
For $\Gamma$ a circle of radius 3 we have no isolated singularites within the loop and so, by Cauchy's theorem, we have
$$\int_\Gamma \frac{1}{(z+4)^3(z-2i)}dz=2\pi i$$
For $\Gamma$ a circle of radius 5 we have 2 isolated singularites which we traverse once and so $$\int_\Gamma \frac{1}{(z+4)^3(z-2i)}dz=2\pi i (1+4+2i)=10\pi i -4\pi$$
I suppose that your circles are centered at $0$.
If the radius is $1$, then what you did is correct.
If the radius is $3$, then there is one isolated singularity, at $2i$. By Cauchy's integral formula, the integral is equal to$$2\pi i\times\frac1{(4+2i)^3}=\frac{11+2i}{500}.$$
If the radius is $5$, then there are two isolated singularities, at $2i$ and at $-4$. By the residue theorem, the integral is equal to $2\pi i$ times the sum of the residues at $2i$ and at $-4$, which happen to be symmetrical. Therefore, the integral is equal to $0$.