Rather strange Inverse Laplace transform

Question

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!

Answer 1

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 $h(t)= [\cos(t)-2\sin(t)]e^{-2t}$.