$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \mathcal{L}\left[ \dfrac{e^{-t} - 1}{t} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \mathcal{L}\left[ \dfrac{e^{-t}}{t} - \dfrac{1}{t} \right]$
$\mathcal{L}(e^{-t}) = \dfrac{1}{s + 1}$
$\mathcal{L}(1) = \dfrac{1}{s}$
From division by t:
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \left[ \int_s^\infty \dfrac{du}{u + 1} - \int_s^\infty \dfrac{du}{u} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \Big[ \ln (u + 1) - \ln u \Big]_s^\infty$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \left[ \ln \dfrac{u + 1}{u} \right]_s^\infty$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \left[ \ln \dfrac{\infty + 1}{\infty} - \ln \dfrac{s + 1}{s}\right]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \left[ -\ln \dfrac{s + 1}{s} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = -\dfrac{1}{s} \left[ \ln (s + 1) - \ln s \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \Big[ \ln s - \ln (s + 1) \Big]$
$\displaystyle \mathcal{L} \left[ \int_0^t \dfrac{e^{-u} - 1}{u} \, du \right] = \dfrac{1}{s} \left[ \ln \dfrac{s}{s + 1} \right]$ answer
