Find the inverse Laplace transform of $$F(s)=\frac{s^2}{s^3 -1}$$
What I've done is to decompose into partial fractions, so I got:
$$\mathcal{L}^{-1}\{\frac{s^2}{s^3 -1}\}=\mathcal{L}^{-1}\{\frac{1}{3(s-1)}+\frac{2s+1}{3(s^2 +s+1)}\}$$
By the properties of the inverse Laplace transform: $$=\mathcal{L}^{-1}\{ \frac{1}{3(s-1)}\}+\mathcal{L}^{-1}\{\frac{2s+1}{3(s^2 +s+1)}\}=\frac{1}{3}\mathcal{L}^{-1}\{ \frac{1}{(s-1)}\}+\frac{1}{3}\mathcal{L}^{-1}\{\frac{2s+1}{s^2 +s+1}\}$$ Where: $$\mathcal{L^{-1}}\{ \frac{1}{(s-1)}\}=e^{t}$$
and $$\mathcal{L}^{-1}\{\frac{2s+1}{s^2 +s+1}\}=\mathcal{L}^{-1}\{\frac{2s}{s^2 +s+1}\}+\mathcal{L}^{-1}\{\frac{1}{s^2 +s+1}\}$$
How can I finish finding the inverse transformation?
In Mathematica, the solution I've got is $$f(t)=\frac{e^{t}}3{}+\frac{2}{3}e^{-t/2}\cos[\frac{\sqrt{3}t}{2}]$$
$s^3-1$ vanishes at the third roots of unity, $1,\omega,\omega^2$. It follows that $$ \frac{s^2}{s^3-1} = \frac{A}{s-1}+\frac{B}{s-\omega}+\frac{C}{s-\omega^2}\tag{1} $$ and by residues: $$ A = \text{Res}\left(\frac{s^2}{s^3-1},s=1\right) = \lim_{s\to 1}\frac{s^2}{s^2+s+1}=\frac{1}{3}, $$ $$ B = \text{Res}\left(\frac{s^2}{s^3-1},s=\omega\right) = \lim_{s\to \omega}\frac{s^2}{(s-1)(s-\omega^2)}=\frac{1}{3}, $$ $$ C = \text{Res}\left(\frac{s^2}{s^3-1},s=\omega^2\right) = \lim_{s\to \omega^2}\frac{s^2}{(s-1)(s-\omega)}=\frac{1}{3}, \tag{2}$$ then since $\mathcal{L}^{-1}\left(\frac{1}{s-s_0}\right)(x)=e^{s_0 x}$ we have: $$ \mathcal{L}^{-1}\left(\frac{s^2}{s^3-1}\right) = \frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3} = \color{red}{\frac{1}{3}e^x+\frac{2}{3}e^{-x/2}\cos\left(\frac{x\sqrt{3}}{2}\right)}\tag{3}$$ as wanted.