Answer
$$\dfrac{6}{s^3}-\dfrac{5s}{s^2-4}+\dfrac{3}{s^2-9}$$
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
We are given that $f(t)=3 t^2 -5 \cos 2t +\sin (3t)$
Now, $L[F(t)]=L[3 t^2 -5 \cos 2t +\sin (3t)] \\=L[3t^2]-L[5 \cos 2t]+L[\sin (3t)] \\=\dfrac{(3)(2)}{s^3}-\dfrac{(5)(s)}{s^2+4}+\dfrac{3}{s^2+9}\\=\dfrac{6}{s^3}-\dfrac{5s}{s^2-4}+\dfrac{3}{s^2-9}$