Use the transform rule $\mathcal{L}^{-1}\{\frac{F(s)}{s}\} = \int_0^t{f(\tau)d\tau}$ to find the inverse transform of $F(s) = \frac{6}{s(s+3)(s+3)}$ How does the transform rule help us solve this problem?  Does this just mean I can rewrite the problem as:
$$\mathcal{L}^{-1}\left\{\frac{6}{(s+3)(s+3)}\right\} = \int_0^t \frac{6}{\tau(\tau+3)(\tau+3)}d\tau$$
Which I can then do what with?

$$ F(s) = \frac{1}{s} \frac{6}{(s+3)^2}  $$ 
$$  \frac{1}{s} \triangleq  \int_0^t $$
$$ f(t)=\int_0^t \mathcal{L^{-1}}[\frac{6}{(s+3)^2}]d\tau  $$
We know :
$$ \mathcal{L}[t^n e^{-at}]=\frac{n!}{(s+a)^{n+1}} $$ 
$$ \mathcal{L^{-1}}[\frac{6}{(s+3)^2}]=6te^{-3t} $$
So : 
$$  f(t)= \int_0^t 6\tau e^{-3 \tau }  d \tau =6[\frac{-t}{3}e^{-3t}-\frac{1}{9}e^{-3t}]|_0^t $$
$$ f(t)=-2te^{-3t}-\frac{2}{3}e^{-3t}+\frac{2}{3} $$
