Answer
$$\dfrac{s-2}{s^2-4s+13}$$
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 $f(t)=e^{2t} \cos (3t)$
Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\=\int_{0}^{\infty} e^{-st} [e^{2t} \cos (3t)] dt\\=\int_{0}^{\infty} e^{-(s-2)t} \cos (3t) \ dt \\=\dfrac{s-2}{s^2-4s+13}$
