inverse laplace transform of $\dfrac{5s^2 - 15s - 11}{(s + 1)(s - 3)^3}$

$\dfrac{5s^2 - 15s - 11}{(s + 1)(s - 3)^3} = \dfrac{A}{s + 1} + \dfrac{B}{s - 3} + \dfrac{C}{(s - 3)^2} + \dfrac{D}{(s - 3)^3}$

$5s^2 - 15s - 11 = A(s - 3)^3 + B(s + 1)(s - 3)^2 + C(s + 1)(s - 3) + D(s + 1)$

$\begin{align} 5s^2 - 15s - 11 = & \, A(s^3 - 9s^2 + 27s - 27) + B(s^3 - 5s^2 + 3s + 9) \\ & + C(s^2 - 2s - 3) + D(s + 1) \end{align}$
 

$\begin{align} \mathcal{L}^{-1} \left[ \dfrac{5s^2- 15s - 11}{(s + 1)(s - 3)^3} \right] & = \mathcal{L}^{-1} \left[ \dfrac{-9/64}{s + 1} + \dfrac{5/64}{s - 3} + \dfrac{33/8}{(s - 3)^2} + \dfrac{-11/4}{(s - 3)^3} \right] \\ & = -\dfrac{9}{64} e^{-t} + \dfrac{5}{64} e^{3t} + \dfrac{33}{8} \dfrac{e^{3t}t^{2 - 1}}{(2 - 1)!} - \dfrac{11}{4} \dfrac{e^{3t}t^{3 - 1}}{(3 - 1)!} \\ & = -\dfrac{9e^{-t}}{64} + \dfrac{5e^{3t}}{64} + \dfrac{33e^{3t}t}{8} - \dfrac{11e^{3t}t^2}{8} \end{align}$