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?

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.
