Can't find the inverse of Laplace Transform. Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}
I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.
Anyone can give me hint to give me hint for this problem?

$$F(s)=\dfrac{2}{s(s^2+4)}$$
$$F(s)=\dfrac{1}s\dfrac 2{(s^2+4)}= \mathcal {L}(1) \mathcal{L} ( \sin (2t)$$
You can also use the theorem of convolution :
$$f(t)=1* \sin (2t)=\int_0^t 1 \times \sin(2\tau) d\tau$$
$$f(t)= \dfrac {-\cos (2 \tau)}{2} \bigg |_0^t$$
$$f(t)= \dfrac 12 -\dfrac {\cos (2 t)}{2} $$
