This is just a demonstration on how to solve the following type of problem.
Find $\mathcal{L}^{-1}\{\frac{54}{s^3(s-3)}\}$ by the given method:
$$\mathcal{L}\{ \int_0^tf(\tau)d\tau\}=\frac{1}{s}\mathcal{L}\{f(t)\}$$
The textbook answer is $f(t)=2e^{3t}-9t^2-6t-2$.
My solution:
$$\frac{54}{s^2(s-3)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-3}$$
$$As(s-3)+B(s-3)+Cs^2=54$$
$$As^2-3As+Bs-3B+Cs^2=54$$
$$A+C=0, -3A+B=0, -3B=54$$
$$A=-6, B=-18, C=6$$
$$f(\tau)=-6-18\tau+6e^{3\tau}$$
$$\mathcal{L}^{-1}\{\frac{1}{s}F(s)\}=\int_0^t(-6-18\tau+6e^{3\tau})d\tau=-6\tau-9\tau^2+2e^{3\tau}\biggr|^{t}_0 $$
$$f(t)=-6t-9t^2+2e^{3t}-2$$