Answer
$1+\sqrt{2}-2^{\frac{3}{4}}$
Work Step by Step
\begin{aligned} \int_{1}^{\sqrt{2}} \frac{s^{2}+\sqrt{s}}{s^{2}} d s&= \int_{1}^{\sqrt{2}}\left(1+ s^{-3/2} \right)d s\\
&= \left(s-2s^{-1/2} \right)\bigg|_{1}^{\sqrt{2}} \\
&= (\sqrt{2}-2(2)^{-1/4}+1)\\
&=1+\sqrt{2}-2^{\frac{3}{4}}
\end{aligned}
