Answer
$$\frac{49}{15}$$
Work Step by Step
Given $$\int_{0}^{1}(\sqrt[4]{u}+1)^2du$$
Since
\begin{align*}
\int_{0}^{1}(\sqrt[4]{u}+1)^2du&=\int_{0}^{1}( u^{1/2}+2u^{1/4}+1) du\\
&=\frac{2}{3}u^{3/2}+\frac{8}{5}u^{5/4}+u\bigg|_{0}^{1}\\
&=\frac{49}{15}
\end{align*}
