The correct answer is $\boxed{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)} + {{s + 2} \over {{s^2} + 1}}$.
The Laplace transform of the product of two functions $f(t)$ and $g(t)$ is given by the convolution theorem:
$$L\left[ {f\left( t \right)g\left( t \right)} \right] = {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} {F\left( s \right)G\left( {s – z} \right)dz}$$
where $F(s)$ and $G(s)$ are the Laplace transforms of $f(t)$ and $g(t)$, respectively, and $c$ is a real number greater than the real parts of all the poles of $F(s)$ and $G(s)$.
In this case, we have:
$$L\left[ {f\left( t \right)g\left( t \right)} \right] = {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} {{s + 2} \over {{s^2} + 1}} \cdot {{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}dz}$$
We can simplify this integral by using partial fractions:
$$\begin{align}
L\left[ {f\left( t \right)g\left( t \right)} \right] &= {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} {{s + 2} \over {{s^2} + 1}} \cdot {{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}dz \\
&= {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} \left( {{1 \over {s + 3}} + {1 \over {s + 2}}} \right) \cdot {{s^2} + 1} dz \\
&= {1 \over {2\pi i}} \left( {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} {{1 \over {s + 3}}dz} + {1 \over {2\pi i}} \int_{c-i\infty}^{c+i\infty} {{1 \over {s + 2}}dz} \right) \cdot {{s^2} + 1} \\
&= {1 \over {2}} \left( {1 \over {s + 3}} \cdot {{s^2} + 1} + {1 \over {s + 2}} \cdot {{s^2} + 1} \right) \\
&= {{s^2} + 1} \over {2\left( {s + 3} \right)\left( {s + 2} \right)} + {{s + 2} \over {2\left( {s^2} + 1} \right)}} \\
&= {{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}} + {{s + 2} \over {{s^2} + 1}}
\end{align}$$
Therefore, the Laplace transform of $h(t)$ is $\boxed{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}} + {{s + 2} \over {{s^2} + 1}}$.