$$L[\sin3u] = \frac{3}{s^2+3^2}$$

$\therefore L[\sin3u] = (-1)\frac{d}{ds}\bigg[\frac{3}{s^2 + 9}\bigg]$

By multiplication of t property

$\therefore L[\sin3u] = (-1)\bigg[\frac{(s^2 + 9)\frac{d}{ds}(3) - 3\frac{d}{ds}(s^2 + 9)}{(s^2 + 9)^2}\bigg] \\ \therefore L[\sin3u] = (-1) \bigg[\frac{-3(2s)}{(s^2 + 9)^2}\bigg] \\ \therefore L[\sin3u] = \frac{6s}{(s^2 + 9)^2} \\ \therefore L\bigg[\int\limits_{0}^{t}u \sin3u du\bigg] = \frac{1}{s}\frac{6s}{(s^2 + 9)^2}$

By applying integral property

$\therefore L\bigg[e^{-3t}\int\limits_{0}^{t} u \sin3udu\bigg] = \frac{6}{[(s+ 3)^2 + 9]^2}$

By first shifting property

$\therefore L\bigg[e^{-3t}\int\limits_{0}^{t} u \sin3udu\bigg] = \frac{6}{(s^2 + 6s + 18)^2}$