Inverse Laplace Transform help Is the information below correct?
Find the inverse Laplace transform of $$ F(s) = \frac{s}{s^2 + 4s + 13}$$
Soln:
a) Complete the squares to simplify our denominator 
$$ s^2 + 4s + 13 = (s+2)^2 + 9 = (s+2)^2 + 3^2$$
$$\mathscr{L}^{-1}\left\{F(s)\right\} = \frac{s}{(s+2)^2 + 3^2}. $$ From the table we can deduce that this is 
$$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$

HINT:
$$
\frac{s}{{(s + 2)^2  + 3^2 }} = \frac{{s + 2 - 2}}{{(s + 2)^2  + 3^2 }} = \frac{{s + 2}}{{(s + 2)^2  + 3^2 }} - \frac{2}{3}\frac{3}{{(s + 2)^2  + 3^2 }}.
$$
Now note that the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2  + \omega ^2 }}$ and  $\frac{\omega}{{(s + \alpha)^2  + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively.
