Given, f(u) = $\int\limits_{0}^{t} ue^{-3u} sin4udu$

We shall first find L[sin(4u)]

$\therefore$ L[sin(4u)] = $\frac{4}{s^2 + 4^2}$ = $\frac{4}{s^2 + 16}$

L[e$^ {-3u}$sin4u] = $\frac{4}{(s+3)^2 + 16}$ [By first shifting property]

[First shifting property L[$e^{at} f(t)$] = $f(s - a)$]

Now by theorem of multiplication of 't', here in this case 'u' we have

L[$ue^{-3u} sin4u$] = (-1) $\frac{d}{ds} \frac{4}{(s + 3)^2 + 16}$

$\therefore$ L[$ue^{-3u} sin4u$] = (-1) $\frac{d}{ds} \frac{4}{s^2 + 6s + 25}$

$\therefore$ L[$ue^{-3u} sin4u$] = (-1) {$ \frac{s^2 + 6s + 25(0) - 4(2s + 6)} {(s^2 + 6s + 25)^2}$}

$\therefore$ L[$ue^{-3u} sin4u$] = (-1) {$\frac{-8(s + 3)}{(s^2 + 6s + 25)^2}$}

$\therefore$ L[$ue^{-3u} sin4u$] = $\frac{8(s+3)}{(s^2 + 6s + 25)^2}$

Hence, finally by theorem of Laplace transform integral.

We have

L[$\int\limits_{0}^{t} ue^{-3u} sin4u du$] = $\frac{8(s+3)}{(s^2 + 6s + 25)^2}$

$\because$ L[$\int\limits_{0}^{t} f(t) dt$] = $\frac{1}{s}F(s)$

$\therefore$ L[$\int\limits_{0}^{t} ue^{-3u} sin4u du$] = $\frac{1}{s}${$\frac{8(s+3)}{(s^2 + 6s + 25)^2}$}