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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?

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Well, can you actually solve the integral and come up with the inverse Laplace transform? – Amzoti Dec 11 '12 at 6:56
Wolfram tells me that the integral does not converge. – John Stewley Dec 11 '12 at 6:58
How about working it from another angle first. What is the inverse Laplace transform of F(s)? (Do it the way you know how.) Can you now figure out how to use that to move forward? – Amzoti Dec 11 '12 at 7:00
Should I use a table to look this up? – John Stewley Dec 11 '12 at 7:03
Do you know the method of partial fractions? – Gyu Eun Lee Dec 11 '12 at 7:09

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

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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.

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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, $$\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}=\int_0^t f(\tau)\,d\tau=\int_0^t \tau e^{-3\tau}\,d\tau=-2 e^{-3 t} t-\frac{2 e^{-3 t}}{3}+\frac{2}{3}.$$

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