Answer
$$\mathcal{L}^{-1}\left\{F(s)\right\}=\dfrac{3}{2}\sin(2t) $$
Work Step by Step
We see that $\dfrac{3}{s^2+4}$ is most closely in the form of #5 in Table 6.2.1. Let $a=2$ in that entry. Then we need to do some algebra to get a 2 in the numerator of the Laplace inverse.
\begin{align*}
\mathcal{L}^{-1}\left\{\dfrac{3}{s^2+4}\right\}&=3\cdot \mathcal{L}^{-1}\left\{\dfrac{1}{s^2+4}\right\} \\
&=\dfrac{3}{2}\cdot \mathcal{L}^{-1}\left\{\dfrac{2}{s^2+(2)^2}\right\}\\
&=\boxed{\color{blue}{\dfrac{3}{2}\sin(2t)}}
\end{align*}
