∫ 1 64 1 + x 3 x d x {\displaystyle \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
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= ∫ 1 64 x − 1 / 2 + x 1 3 − 1 2 {\displaystyle \int _{1}^{64}x^{-1/2}+x^{{\frac {1}{3}}-{\frac {1}{2}}}} = ∫ 1 64 x − 1 2 + x − 1 6 {\displaystyle \int _{1}^{64}x^{-{\frac {1}{2}}}+x^{-{\frac {1}{6}}}}
Add one to the exponents and divide by the new exponent
= ∫ 1 64 x 1 2 1 2 + x 5 6 5 6 {\displaystyle \int _{1}^{64}{\frac {x^{\frac {1}{2}}}{\frac {1}{2}}}+{\frac {x^{\frac {5}{6}}}{\frac {5}{6}}}} = ∫ 1 64 2 x 1 2 + 6 5 x 5 6 {\displaystyle \int _{1}^{64}2x^{\frac {1}{2}}+{\frac {6}{5}}x^{\frac {5}{6}}}
= 2 ( x ) 1 2 + 6 5 ( x ) 5 6 | 1 64 {\displaystyle 2(x)^{\frac {1}{2}}+{\frac {6}{5}}(x)^{\frac {5}{6}}{\bigg |}_{1}^{64}} = 2 ( 64 ) 1 2 + 6 5 ( 64 ) 5 6 − ( 2 ( 1 ) 1 2 + 6 5 ( 1 ) 5 6 ) {\displaystyle 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 ) {\displaystyle 16+38.4-(2+1.2)}
= 54.4 − 3.2 {\displaystyle 54.4-3.2} = 51.2 {\displaystyle 51.2}
= 256 5 {\displaystyle {\frac {256}{5}}}
![{\displaystyle \int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cdb02981593f318bab17caae838949d55b200cb6)
= \int_{1}^{64}\left(\frac{1}{x^{1/2}} + \frac{x^{1/3}}{x^{1/2}}\right)dx
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