y' = 3\sqrt{x}. 1 + (y')^{2} = 1 + 9x.
\displaystyle s(x) = \int_{1}^{x} \sqrt{1 + 9t} dt = \left[\dfrac{2}{27} (1 + 9t)^{\frac{3}{2}}\right]_{1}^{x}
\displaystyle s(x) = \dfrac{2}{27} (1 + 9x)^{\frac{3}{2}} - \dfrac{2}{27} (10)^{\frac{3}{2}} = \dfrac{2}{27} (1 + 9x)^{\frac{3}{2}} - \dfrac{20\sqrt{10}}{27}