Answer
$$\dfrac{2}{s+3}+\dfrac{4}{(s-1)}-\dfrac{5}{s^2+1}$$
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 $f(t)=2 e^{-3t}+4e^t-5 \sin t$
Now, $L[F(t)]= L[2 e^{-3t}+4e^t-5 \sin t] \\= L[2 e^{-3t}] +L[4e^{t}t]-L[5 \sin t]\\=\dfrac{(2)}{s+3}+\dfrac{4}{(s-1)}-\dfrac{5}{s^2+1}\\=\dfrac{2}{s+3}+\dfrac{4}{(s-1)}-\dfrac{5}{s^2+1}$
