The above equation is of type $\int\limits_{0}^{\infty}e^{-3t}tcostdt$ where s = 3

Taking Laplace of tcost i.e. f(t) and substitutitng s = 3

$\therefore L[cost] = \frac{s}{s^2 + 1}$

$\therefore L[tcost] = -1\frac{d}{ds}\bigg[\frac{s}{s^2 + 1}\bigg]$ By multiplication by t rule

$\therefore L[tcost] = -1\bigg[\frac{(s^2 + 1)\frac{d}{ds}s - s\frac{d}{ds}(s^2 + 1)}{(s^2 + 1)^2}\bigg]$

$\therefore L[tcost] = -1\bigg[\frac{(s^2 + 1) - 2s^2}{(s^2 + 1)^2}\bigg]$

$\therefore L[tcost] = -1\frac{(1 - s^2)}{(s^2 + 1)^2}$

$\therefore L[tcost] = \frac{s^2 - 1}{(s^2 + 1)^2}$

Now s = 3 we get

$\int\limits_{0}^{\infty}e^{-3t}tcostdt = \frac{8}{100} = \frac{2}{25}$