Answer
$\mathcal{L}^{-1}[F(s)]=\dfrac{6}{5} (e^{-2t})+\dfrac{9}{5} (e^{3t})$
Work Step by Step
Rewrite the given function so that it can be written into the partial fraction decomposition.
$\dfrac{3s}{s^2-s-6}=\dfrac{3s}{(s+2)(s-3)}$
and $\dfrac{3s}{(s+2)(s-3)}=\dfrac{A}{(s+2)}+\dfrac{B}{(s-3)}$
Our next step will be to multiply both sides to solve for $A$ and $B$.
$3s=A(s-3)+B(s+2) \implies A=\dfrac{2}{3}B$
Also,we have $A+B=3 \implies \dfrac{2}{3}B+B=3$
This implies that $A=\dfrac{6}{5}$ and $B=\dfrac{9}{5}$
Thus, we get $\dfrac{3s}{(s+2)(s-3)}=\dfrac{6/5}{(s+2)}+\dfrac{9/5}{(s-3)}$
Since, $\mathcal{L}(e^{at})=\dfrac{1}{s-a}$
Therefore, the required Laplace transform is:
$\dfrac{6/5}{(s+2)}+\dfrac{9/5}{(s-3)}=\mathcal{L}(\dfrac{6}{5} e^{-2t})+\mathcal{L}(\dfrac{9}{5} e^{3t})$
Hence, $\mathcal{L}^{-1}[F(s)]=\dfrac{6}{5} (e^{-2t})+\dfrac{9}{5} (e^{3t})$
