The integral is the following:
$$\int_{|z|=r} \frac{z+1}{z(z^2+4)} dz , r>0, r \neq 2 $$
I'm a little bit lost, I know that its partial fraction expansion is
$$ \frac{z+1}{z(z^2+4)} = \frac{4-z}{4(z^2+4)} + \frac{1}{4z} $$ but this doesn't seem like a Laurent series expansion and I don't know from here what is supposed to be the residue, so I can calculate the integral by residues.
Put
$$f(z)=\frac{z+1}{z(z^2+4)}$$
so
$$\begin{align*}&\bullet\;\;\;\text{Res}_{z=0}(f)=\lim_{z\to 0}zf(z)=\frac14\\ &\bullet\;\;\;\text{Res}_{z=\pm 2i}(f)=\lim_{z\to\pm2i}(z\pm2i)f(z)=\frac{1\pm2i}{\pm2i(4i)}=\pm\frac18(1\pm2i)\end{align*}$$ $${}$$
Now use the Residue theorem for integrals, depending on whether $\;0<r<2\;$ , or $\;r>2\;$