So, we need to calculate residue in each pole of
$$\frac{z+1}{(z^2+4)(z-1)^3}.
$$
Points $\pm 2i$ are easy since they are simple poles, with the residues equal to
$$
\frac{z+1}{(z^2+4)'(z-1)^3}\Bigr|_{z=\pm 2i}=\frac{z+1}{2z (z-1)^3}\Bigr|_{z=\pm 2i}=\frac{\pm 2i+1}{\pm 4i (\pm 2i-1)^3}=...
$$
Point $z=1$ is more tricky since it is a triple pole but the factor $(z-1)^3$ is already separated, so we need to find a coefficient at $(z-1)^2$ in the decomposition of $g(z)=\frac{z+1}{(z^2+4)}$ at $z=1$; it is $\frac{1}{2}g''(z)$ at that point.
But there is a simpler way to find a residue at $z=1$. The function is meromorphic in the extended complex plane, having only isolated points. Then the sum of all residues should be $0$, and we need to include $\infty $ in the tally. But since at infinity the function decays faster than $z^{-1}$, the residue there is $0$. So
$\newcommand{\Res}{\operatorname{Res}}$
$$
\Res (f, 1)= -\Res (f,2i) -\Res (f, -2i).
$$
