Using partial fractions, find the inverse Laplace transform of the function $$f(s) = \frac{s^2+2}{s^2(s^2+3)}$$
I have tried to separate the whole function using a partial fraction via
$$\begin{align} \frac{s^2+2}{s^2(s^2+3)} &= \frac{A}{s^2} + \frac{B}{s^2+3} \\ s^2+2 &= A(s^2+3) + B(s^2) \\ s &= \pm\sqrt3, 0 \end{align}$$
I am kind of stuck after this. Am I on the right track?
\begin{align} \frac{s^2+2}{s^2(s^2+3)}&=\frac{A}{s} +\frac{B}{s^2} + \frac{C}{s^2+3}\\ &=\frac{As(s^2+3)+B(s^2+3)+Cs^2}{s^2(s^2+3)}\\ &=\frac{As^3+(B+C)s^2+3As+3B}{s^2(s^2+3)}\end{align} So equating the powers of $s$ you have $A=0$, $3B=2$ and $B+C=1$ that is $B=2/3$ and $ C=1/3$ and then $$f(s)=\frac{2}{3 s^2}+\frac{1}{3 (s^2+3)}$$ So you have $$ \mathcal L\{f(s)\}=\left[\frac{2}{3}t+\frac{1}{3\sqrt 3}\sin(\sqrt 3 t)\right]u(t) $$