QuestionThe integral $\int_{0}^{\infty} e^{-s t} e^{3 t} \cos (2 t) d t=$
Select one: $\frac{3!}{s^{2}-9}$ $\frac{s-3}{(s-3)^{2}+4}$ $\frac{s}{(s+3)^{2}-4}$ $\frac{s}{s^{2}+4}$

Studdy Solution
Simplify the expression:
$\frac{3-s}{(3-s)^2 + 4} = \frac{s-3}{(s-3)^2 + 4}$
This matches the option:
$\boxed{\frac{s-3}{(s-3)^2 + 4}}$
The value of the integral is:
$\boxed{\frac{s-3}{(s-3)^2 + 4}}$