$\mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \mathcal{L} \left[ \dfrac{e^{-3t}}{t} \right] - \mathcal{L} \left[ \dfrac{e^{-6t}}{t} \right]$
Since
$\mathcal{L}(e^{-3t}) = \dfrac{1}{s + 3}$ and
$\mathcal{L}(e^{-6t}) = \dfrac{1}{s + 6}$
Then,
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \int_0^\infty \dfrac{du}{u + 3} - \int_0^\infty \dfrac{du}{u + 6}$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \Big[ \ln (u + 3) \Big]_0^\infty - \Big[ \ln (u + 6) \Big]_0^\infty$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \Big[ \ln \infty - \ln (0 + 3) \Big] - \Big[ \ln \infty - \ln (0 + 6) \Big]$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \Big[ \ln \infty - \ln 3 \Big] - \Big[ \ln \infty - \ln 6 \Big]$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \ln 6 - \ln 3$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \ln (6 / 3)$
$\displaystyle \mathcal{L} \left[ \dfrac{e^{-3t} - e^{-6t}}{t} \right] = \ln 2$
Therefore,
$\displaystyle \int_0^\infty \dfrac{e^{-3t} - e^{-6t}}{t} ~dt = \ln 2$ answer
