Calculus 8th Edition

$\frac{2}{27}[(1+9x)^{3/2} - 10\sqrt{10}]$
$y = 2x^{3/2}$ then $y' = 3x^{1/2}$ and $1+(y')^{2} = 1+9x$ Starting at $P_0 (1, 2)$ $s(x) = \int^{x}_{1} \sqrt{1+9t} dt = [\frac{2}{27} (1+9t)^{3/2}]^{x}_{1} = \frac{2}{27}[(1+9x)^{3/2} - 10\sqrt{10}]$