Problem 02
Find the Laplace transform of $\displaystyle \int_0^t (u^2 - u + e^{u}) \, du$.
Solution 02
$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
Hence,
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{\mathcal{L}(t^2 - t + e^t)}{s}$
$\displaystyle \mathcal{L} \left[ \int_0^t (u^2 - u + e^{u}) \, du \right] = \dfrac{1}{s} \mathcal{L}(t^2 - t + e^t)$