Evaluate the integral $\int_{1}^{\sqrt{2}} \frac{s^{2}+\sqrt{s}}{s^{2}} d s$

Step 1 ：Given the integral \(\int_{1}^{\sqrt{2}} \frac{s^{2}+\sqrt{s}}{s^{2}} d s\)

Step 2 ：Simplify the integral by dividing each term in the numerator by \(s^{2}\), resulting in \(\int_{1}^{\sqrt{2}} (1 + \frac{1}{\sqrt{s}}) ds\)

Step 3 ：Break the integral into two simpler integrals: \(\int_{1}^{\sqrt{2}} ds + \int_{1}^{\sqrt{2}} \frac{1}{\sqrt{s}} ds\)

Step 4 ：Evaluate the first integral to get \([\sqrt{2} - 1]\)

Step 5 ：Evaluate the second integral to get \([2\sqrt{s}]_{1}^{\sqrt{2}}\)

Step 6 ：Add these two results together to get the final answer: \(-3 + \sqrt{2} + 2\cdot2^{1/4}\)

Step 7 ：Final Answer: \(\boxed{-3 + \sqrt{2} + 2\cdot2^{1/4}}\)