Find the inverse Laplace transform of $e^{-3s} \frac {3s+1}{s^2-s-6}$ Problem: Find the inverse Laplace transform of $e^{-3s} \frac {3s+1}{s^2-s-6}$
My attempt:
Using the method of partial fractions $\frac {3s+1}{s^2-s-6}=\frac {2}{s-3} + \frac {1}{s+2}$, so that $L^{-1}(e^{-3s}\frac {3s+1}{s^2-s-6})=2L^{-1}(e^{-3s}\frac {1}{s-3}) + L^{-1}(e^{-3s}\frac {1}{s+2})$.
I am using the second shifting theorem but I am unsure if I am on the right track, the answer I get using my understand of the second shifting theorem is:
$u(t-3)(2e^{(3-3)t}+e^{(-2-3)(t)}) = u(t-3)(2+e^{-5t})$ where $u(t)$ is the heaviside function. 

$$f(t)=\mathcal{L^{-1}}( e^{-3s} \frac {3s+1}{s^2-s-6})=\mathcal{L^{-1}}( e^{-3s} \frac {3s+1}{(s+2)(s-3)})$$
$$f(t)=\mathcal{L^{-1}}( e^{-3s} \frac {2s+4+s-3}{(s+2)(s-3)})$$
$$f(t)=\mathcal{L^{-1}}( e^{-3s} \left (\frac {2}{(s-3)}+\frac {1}{(s+2)} \right ))$$
$$f(t)=2\mathcal{L^{-1}} \left (e^{-3s}\frac {1}{s-3}\right )+\mathcal{L^{-1}} \left ( e^{-3s}\frac {1}{(s+2)} \right )$$
$$f(t)=2U(t-3)e^{3(t-3)}+U(t-3)e^{-2(t-3)}$$
Note that you have that
$$\boxed {\mathcal{L^{-1}} \left ( e^{-cs}F(s) \right )=U(t-c)f(t-c)}$$
