Answer
$\displaystyle \frac{69}{4}=17.25$
Work Step by Step
$\displaystyle \int_1^8\frac{2+t}{t^{2/3}}dt = \int_1^8\frac{2}{t^{2/3}}+\frac{t}{t^{2/3}}dt=\int_1^82t^{-2/3}+t^{1/3}dt$
$\displaystyle (\frac{2t^{1/3}}{1/3}+\frac{t^{4/3}}{4/3})|_1^8$
$\displaystyle (6t^{1/3}+\frac{3}{4}t^{4/3})|_1^8$
$\displaystyle [6(8)^{1/3}+\frac{3}{4}(8)^{4/3}]-[6(1)^{1/3}+\frac{3}{4}(1)^{4/3}]$
$\displaystyle [6(2)+\frac{3}{4}(2)^4]-[6+\frac{3}{4}]$
$\displaystyle 12+\frac{3}{4}(16)]-6-\frac{3}{4}$
$\displaystyle 12-6-\frac{3}{4}+3(4)$
$\displaystyle 6+12-\frac{3}{4}$
$\displaystyle 18-\frac{3}{4}$
$\displaystyle \frac{72}{4}-\frac{3}{4}$
$\displaystyle \frac{69}{4}=17.25$
