Answer
$\sqrt{2}-\sqrt[4] 8+1$
Work Step by Step
$\int^{\sqrt{2}}_{1}\frac{s^2+\sqrt{s}}{s^2}ds$
=$\int^{\sqrt{2}}_{1}(1+s^{\frac{-3}{2}})ds$
=$[s-\frac{2}{\sqrt{s}}]_1^{\sqrt{2}}$
=$(\sqrt{2}-\frac{2}{\sqrt{\sqrt{2}}})-(1-\frac{2}{\sqrt{1}})$
=$\sqrt{2}-2^\frac{3}{4}+1$
=$\sqrt{2}-\sqrt[4] 8+1$
