Answer
$\frac{49}{15}$
Work Step by Step
$\int^1_{0}(\sqrt[4] u+1)^2du=\int^1_{0}(\sqrt u+2u^\frac{1}{4}+1)du$
$=[\frac{2}{3}u^\frac{3}{2}+\frac{8}{5}u^\frac{5}{4}+u] ^1_{0}$.
Sub in the bounds and subtract the lower bound (0) from the upper bound (1):
$= [\frac{2}{3}(1)^\frac{3}{2}+\frac{8}{5}(1)^\frac{5}{4}+(1)]-[\frac{2}{3}(0)^\frac{3}{2}+\frac{8}{5}(0)^\frac{5}{4}+(0)]$
$= \frac{2}{3}+\frac{8}{5}+1-0$
$=\frac{49}{15}$.
