It's a situation similar to finding antiderivatives. We need to know some standard inverse Laplace transforms or search them in a table. In this case since $\mathcal{L}^{-1}\left( \frac{s}{s^{2}+a^{2}}\right) =\cos at,$ $\mathcal{L}^{-1}\left( \frac{a}{ s^{2}+a^{2}}\right) =\sin at,$ and the Inverse Laplace Transform is linear, we have
$\begin{eqnarray*} \mathcal{L}^{-1}(F(s)) &=&\mathcal{L}^{-1}\left( \frac{2s+1}{s^{2}+9}\right) =\mathcal{L}^{-1}\left( \frac{2s}{s^{2}+9}\right) +\mathcal{L}^{-1}\left( \frac{1}{s^{2}+9}\right) \\ &=&2\mathcal{L}^{-1}\left( \frac{s}{s^{2}+9}\right) +\frac{1}{3}\mathcal{L} ^{-1}\left( \frac{3}{s^{2}+9}\right) \\ &=&2\cos 3t+\frac{1}{3}\sin 3t. \end{eqnarray*}$
Any rational function $P(s)/Q(s)$, where the degree of the polynomial $P(s)$ is lower than the degree of the polynomial $Q(s)$ can be expanded into partial fractions, each of them having a standard inverse Laplace Transform.
Another possibility is to use the Heaviside expansion formula $ \mathcal{L}^{-1}\left( \frac{P(s)}{Q(s)}\right) =\sum_{k=1}^{n}\frac{P(s_{k})}{Q^{\prime }(s_{k})}e^{s_{k}t}, $ where the $s_{k}$ are the distinct zeroes of $Q(s)$ (Schaum's Outline of Theory and Problems of Laplace Transforms by Murray Spiegel, Portuguese translation).