Inverse Laplace Transform partial fraction $\frac{2}{s(s^2+s)}$

Question

Can anybody help me with the answer of this question?

Find the inverse Laplace transform of: $$\frac{2}{s(s^2+s)}$$

Answer 1

Since $$\frac{2}{s^2(s+1)}=\frac{2}{s}\left(\frac{1}{s}-\frac{1}{s+1}\right)=\frac{2}{s^2}-\frac{2}{s(s+1)}=\frac{2}{s^2}-\frac{2}{s}+\frac{2}{s+1},$$the fact that $e^{ax},\,xe^{ax}$ have respective Laplace transforms $\frac{1}{s-a},\,\frac{1}{(s-a)^2}$ implies the inverse Laplace transform of $\frac{2}{s^2(s+1)}$ is $2x-2+2e^{-x}$.