TITLE: Inverse Laplace Transform help
QUESTION [8 upvotes]: 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 
$$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$

REPLY [3 votes]: For rational functions you can use decomposition to partial fractions:
$$\frac{s}{(s+2)^2+3^2}=\frac{s}{(s+2+3i)(s+2-3i)}=\frac{1/2+i/3}{s+2-3i}+\frac{1/2-i/3}{s+2+3i}.$$
The only thing you need to know now is that the inverse Laplace transform of $1/(s+\alpha)$ is $\exp(-\alpha x)$.