= Solution
A <Laurent series> about $a$ is an expansion
$$
f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n
$$
that converges to $f$ on an <annulus> $r<|z-a|<R$.
The <partial fraction decomposition> is
$$
\frac{10}{(z+2)(z^2+1)}
=\frac2{z+2}+\frac{4-2z}{z^2+1}.
$$
For $0<|z|<1$, expand both denominators by the <geometric series>:
$$
\boxed{f(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^n}z^n
+(4-2z)\sum_{k=0}^{\infty}(-1)^kz^{2k}}.
$$
This is actually a <Taylor series> at zero, since the apparent puncture at zero contains no singularity.
For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:
$$
\frac{4-2z}{z^2+1}
=(4z^{-2}-2z^{-1})\frac1{1+z^{-2}}.
$$
Hence
$$
\boxed{f(z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^n}z^n
+(4z^{-2}-2z^{-1})\sum_{k=0}^{\infty}(-1)^kz^{-2k}}.
$$
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