Answer
$\dfrac{2}{(s-3)^2+1}+\dfrac{4(s+1)}{(s+1)^2+9}$
Work Step by Step
The Laplace transform of function $t^3$ is given as:
$L(2 \sin t)=\dfrac{2}{s^2+1}$ and $L( 4 \cos 3t)=\dfrac{4s}{s^2+9}$
The first shifting Theorem for $a=- 2$ can be expressed as:
$L[2e^{3t} \sin t+4e^{-t} \cos 3t]=\dfrac{2}{(s-3)^2+1}+\dfrac{4(s+1)}{(s+1)^2+9}$
