5.4 Indefinite Integrals and the Net Change Theorem/39

${\begin{aligned}\int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx&=\int _{1}^{64}\left({\frac {1}{x^{1/2}}}+{\frac {x^{1/3}}{x^{1/2}}}\right)dx=\int _{1}^{64}\left(x^{-1/2}+x^{{\frac {1}{3}}-{\frac {1}{2}}}\right)dx=\int _{1}^{64}\left(x^{-{\frac {1}{2}}}+x^{-{\frac {1}{6}}}\right)dx\\[2ex]&=\left[{\frac {x^{\frac {1}{2}}}{\frac {1}{2}}}+{\frac {x^{\frac {5}{6}}}{\frac {5}{6}}}\right]_{1}^{64}=\left[2x^{\frac {1}{2}}+{\frac {6}{5}}x^{\frac {5}{6}}\right]_{1}^{64}\\[2ex]&=2(x)^{\frac {1}{2}}+{\frac {6}{5}}(x)^{\frac {5}{6}}{\bigg |}_{1}^{64}2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}-(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})\end{aligned}}$

= $2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}-(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})$

= $16+38.4-(2+1.2)$

= $54.4-3.2$ = $51.2$

= ${\frac {256}{5}}$