Find the inverse Laplace transform of $\frac{s^2}{s^3 -1}$ 
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.
