Inverse Laplace Transform help

Question

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 $$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$

Answer 1

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.

Answer 2

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 $\exp(-\alpha x)$.