$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$
$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} (1) \, dt$
$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} \, dt$
$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \int_0^\infty e^{-st} \, (-s \, dt)$
$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \left[ e^{-st} \right]_0^\infty$
$\mathcal{L} (1) = -\dfrac{1}{s} \left[ \dfrac{1}{e^{st}} \right]_0^\infty$
$\mathcal{L} (1) = -\dfrac{1}{s} \left[ \dfrac{1}{\infty} - \dfrac{1}{e^0} \right]$
$\mathcal{L} (1) = -\dfrac{1}{s} (0 - 1)$
Thus,
$\mathcal{L} (1) = \dfrac{1}{s}$ answer
