$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
Since,
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$
$F(s) = \arctan \left( \dfrac{1}{s} \right)$
Then,
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{\sin u}{u} \, du \right] = \dfrac{\arctan \left( \dfrac{1}{s} \right)}{s}$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{\sin u}{u} \, du \right] = \dfrac{1}{s}\arctan \left( \dfrac{1}{s} \right)$ answer