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).