Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 14

Find the intergral $\displaystyle \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du$, if it exists.

$\begin{equation} \begin{aligned} \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \int^1_0 \left[ (\sqrt[4]{u} )^2 + 2 \sqrt[4]{u} + 1 \right] du \\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \int^1_0 \left( \sqrt{u} + 2 \sqrt[4]{u} + 1 \right) du \\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \int^1_0 u^{\frac{1}{2}} + 2 u^{\frac{1}{4}} + 1 du\\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \left[ \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2}+ 1} + 2 \left( \frac{u^{\frac{1}{4}+1}}{\frac{1}{4}+1} \right) + u \right]^1_0\\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + 2 \left( \frac{u^{\frac{5}{4}}}{\frac{5}{4}} \right) + u \right]^1_0\\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \left[ \frac{2u^{\frac{3}{2}}}{3} + \frac{8u^{\frac{5}{4}}}{5} + u \right]^1_0\\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \frac{2(1)^{\frac{3}{2}}}{3} + \frac{8(1)^{\frac{5}{4}}}{5} + 1 - \frac{2(0)^{\frac{3}{2}}}{3} - \frac{(0)^{\frac{5}{4}}}{5} - 0 \\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \frac{2}{3} + \frac{8}{5} + 1 \\ \\ \int^1_0 \left( \sqrt[4]{u} + 1 \right)^2 du &= \frac{49}{15} \end{aligned} \end{equation}$