How to find the inverse Laplace transform?

Question

I'm trying to calculate

$$\mathcal{L}^{-1}\left(\frac{3s^3-3s^2+3s-5}{s^2(s^2+2s+5)}\right)$$

But I am not sure how to go from here. I would be really grateful for any help. Thanks.

Answer 1

Hint (@Amzoti's comment, rephrased): Start with $$\mathcal{L}^{-1}\left(\frac1s\right),\quad \mathcal{L}^{-1}\left(\frac1{s^2}\right),\quad \mathcal{L}^{-1}\left(\frac1{s+1+2\mathrm i}\right),\quad \mathcal{L}^{-1}\left(\frac1{s+1-2\mathrm i}\right).$$

Answer 2

We can work directly with the inverse Laplace transform and poles of $s = 0, -1\pm 2i$. $$ \mathcal{L}^{-1}\{F(s)\}(t)=\frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{3s^3-3s^2+3s-5}{s^2(s^2+2s+5)}e^{st}ds=\sum\text{Res}\{F(s);z_j\} $$ Then the sum of residues are \begin{gather} \lim_{s\to 0}\frac{d}{ds}s^2\frac{3s^3-3s^2+3s-5}{s^2(s^2+2s+5)}e^{st}+\lim_{s\to -1+2i}(s+1-2i)\frac{3s^3-3s^2+3s-5}{s^2(s^2+2s+5)}e^{st}+\lim_{s\to -1+2i}(s+1-2i)\frac{3s^3-3s^2+3s-5}{s^2(s^2+2s+5)}e^{st}ds\\ =1-t-\frac{1}{2}e^{(-1+2i)t}-\frac{1}{2}e^{-(1+2i)t} \end{gather} which can be reduced however you see fit.