Answer
See below
Work Step by Step
Given: $F(s)=\frac{1}{s+2}\\
G(s)=\frac{s+2}{s^2+4s+13}$
a) Using the Convolution Theorem
$L^{-1}[F(s)* G(s)]=L^{-1}[\frac{1}{s+2}.\frac{s+2}{s^2+4s+13}]\\
=L^{-1}[\frac{1}{s+2}].L^{-1}[\frac{s+2}{s^2+4s+13}]\\
=e^{-2t}.e^{-2t}\cos 3t\\
=\int^t_0 e^{-2(t-x)}e^{-2x}\cos 3xdx\\
=e^{-2t}\int^t_0 \cos 3x dx\\
=\begin{bmatrix}
\frac{e^{-2t}}{3}\sin 3x
\end{bmatrix}^t_0\\
=\frac{e^{-2t}}{3}\sin 3t$
b) Using partial fractions.
$L^{-1}[F(s) * G(s)]=L^{-1} \begin{bmatrix}
\frac{1}{s+2}.\frac{s+2}{s^2+4s+13}
\end{bmatrix}\\
=L^{-1}\begin{bmatrix}
\frac{1}{s^2+4s+13}
\end{bmatrix}\\
=\frac{1}{3}L^{-1}\begin{bmatrix}
\frac{3}{(s+2)^2+9}
\end{bmatrix}\\
=\frac{e^{-2t}}{2}\sin 3t$
