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?
 A: $$ F(s) = \frac{1}{s} \frac{6}{(s+3)^2}  $$ 
$$  \frac{1}{s} \triangleq  \int_0^t $$
$$ f(t)=\int_0^t \mathcal{L^{-1}}[\frac{6}{(s+3)^2}]d\tau  $$
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 : 
$$  f(t)= \int_0^t 6\tau e^{-3 \tau }  d \tau =6[\frac{-t}{3}e^{-3t}-\frac{1}{9}e^{-3t}]|_0^t $$
$$ f(t)=-2te^{-3t}-\frac{2}{3}e^{-3t}+\frac{2}{3} $$
A: Hint:
If $A,B,C$ are constants, then given the rational function
$\displaystyle f(x) = \frac{A}{(x-B)(x-C)}$
we can write
$\displaystyle f(x)=\frac{\alpha}{x-B}+\frac{\beta}{x-C}$ for some constants $\alpha, \beta$. This in fact generalizes to when
$\displaystyle f(x) = \frac{A}{(x-B)(x-C)(x-D)}$, and when
$\displaystyle f(x) = \frac{A}{(x-B)(x-C)(x-D)(x-E)}$, and so on, and when the terms in the denominator repeat.
If you know how to get $\alpha$ and $\beta$, then the rest of the problem will be easily solved.
A: Amir Alizadeh has all the pieces, I'm just tying it together/explaining it slightly differently.
You want $\displaystyle{\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}}$ and this matches the form of your hint with $F(s)={6\over (s+3)^2}$ which implies $f(t)=6te^{-3t}$. Thus, $$\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}=\int_0^t f(\tau)\,d\tau=\int_0^t \tau e^{-3\tau}\,d\tau=-2 e^{-3 t} t-\frac{2 e^{-3 t}}{3}+\frac{2}{3}.$$
