Find the Inverse Laplace Transforms Find the inverse Laplace transform of:
$$\frac{3s+5}{s(s^2+9)}$$
Workings:
$\frac{3s+5}{s(s^2+9)}$
$= \frac{3s}{s(s^2+9} + \frac{5}{s(s^2+9)}$
$ = \frac{3}{s^2+9} + \frac{5}{s}\frac{1}{s^2+9}$
$ = \sin(3t) + \frac{5}{s}\frac{1}{s^2+9}$
Now I'm not to sure on what to do.
Any help will be appreciated.

The convolution theorem is, found here,
\begin{align}
\mathcal{L}^{-1} \{f(s)g(s)\} = \int_{0}^{t} f(t-u) g(u) \, du
\end{align}
In the case here $f(s) = 1/s$ which is the transform of $1$ and $g(s)$ being the transform of $\sin$. Now
\begin{align}
\mathcal{L}^{-1}\{ \frac{1}{s ( s^{2} + a^{2})} \} &= \frac{1}{a} \, \int_{0}^{t} (1) \, \sin(a u) \, du = - \frac{1}{a} \, \left[ \frac{\cos(au)}{a} \right]_{0}^{t} \\
&= - \frac{1}{a^{2}} ( \cos(at) - 1)
\end{align} 
With all this it is seen that:
\begin{align}
\mathcal{L}^{-1}\{ \frac{3 s + 5}{ s (s^{2} + 9) } \} &= \mathcal{L}^{-1}\{ \frac{3}{ s^{2} + 9 } \} + \mathcal{L}^{-1}\{ \frac{5}{ s (s^{2} + 9) } \} \\
&= \sin(3t) + \frac{5(1 - \cos(3t))}{9}
\end{align}
