Answer
$\frac{2}{243} (82 \sqrt{82} - 1)$
Work Step by Step
$y = 1+6x^{3/2}$ then $dy/dx = 9x^{1/2}$ and $1+(dy/dx)^{2} = 1+81x$
$L = \int^{1}_{0} \sqrt{1+81x} dx = \int ^{82}_{1} u^{1/2}(\frac{1}{81}du) = \frac{1}{81} \frac{2}{3} [u^{3/2}]^{82}_{1} = \frac{2}{243} (82 \sqrt{82} - 1)$
