Rather strange Inverse Laplace transform I want to find the inverse Laplace transform of:
\begin{equation}
\frac{s}{s^2+4s+5}
\end{equation}
So I rewrite it to:
\begin{equation}
\frac{s}{(s+2)^2+1^2}
\end{equation}
I see that I can extract $s+2 = p$, and thus add $+2$ and $-2$ to the numerator:
\begin{equation}
\frac{(s+2)}{(s+2)^2+1^2}-\frac{2}{(s+2)^2+1^2}
\end{equation}
The first form, with $p=s+2$ gives  $\mathscr{L}^{-1}\frac{p}{p^2+1^2}=\cos t$. The second, with $p=s+2$, $\mathscr{L}^{-1}\frac{2}{p^2+1^2}=\sin(2t)$, so I would be tempted to write:
\begin{equation}
\mathscr{L}^{-1}\frac{s}{s^2+4s+5}=\cos t+\sin(2t)
\end{equation}
But this is wrong! So how can I get this right?
Thanks!

Note that
$$
H(s)
=
\frac{s}{s^2+4s+5}
=
\frac{(s+2)-2}{(s+2)^2+1^2}
=
G_1(s+2)- 2G_2(s+2)
$$
where
$$
G_1(s) = \frac{s}{s^2+1^2},
G_2(s) = \frac{1}{s^2+1^2}
$$
We deduce
$h(t)=
[\cos(t)-2\sin(t)]e^{-2t}$.
