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**skeeter** $\displaystyle f'(x) = \frac{2}{3}x^{-\frac{1}{3}}$

$\displaystyle f'(x) = \frac{2}{3x^{\frac{1}{3}}}$

$\displaystyle [f'(x)]^2 = \frac{4}{9x^{\frac{2}{3}}}$

$\displaystyle 1 + [f'(x)]^2 = 1 + \frac{4}{9x^{\frac{2}{3}}} = \frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}

$

$\displaystyle \sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4}$

$\displaystyle \int_1^8 \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4} \, dx$

let $\displaystyle u = 9x^{\frac{2}{3}}+4$

$\displaystyle du = \frac{6}{x^{\frac{1}{3}}} \, dx$

$\displaystyle \frac{1}{18} \int_{13}^{40} \sqrt{u} \, du$

can you finish ?