5.4 Indefinite Integrals and the Net Change Theorem/39

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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]&=\left[2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}\right]-\left[(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})\right]\\[2ex]&=\left[2\cdot 8+{\frac {6}{5}}(2)^{5}\right]-\left[2+{\frac {6}{5}}\right]=\left[16+{\frac {192}{5}}\right]-\left[{\frac {16}{5}}\right]=\left[{\frac {80}{5}}+{\frac {192}{5}}\right]-\left[{\frac {16}{5}}\right]\\[2ex]&={\frac {256}{5}}\end{aligned}}}