$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
Hence,
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{\mathcal{L}(t^2 - t + e^t)}{s}$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \mathcal{L}(t^2 - t + e^t)$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \left[ \dfrac{2}{s^3} - \dfrac{1}{s^2} + \dfrac{1}{s - 1} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \left[ \dfrac{2(s - 1) - s(s - 1) + s^3}{s^3(s - 1)} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \left[ \dfrac{2s - 2 - s^2 + s + s^3}{s^3(s - 1)} \right]$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{s^3 - s^2 + 3s - 2}{s^4(s - 1)}$ answer
