∫ 1 64 1 + x 3 x d x {\displaystyle \int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx}
= ∫ 1 64 1 x 1 / 2 {\displaystyle \int _{1}^{64}{\frac {1}{x^{1/2}}}} + ∫ 1 64 x 1 / 3 x 1 / 2 {\displaystyle \int _{1}^{64}{\frac {x^{1/3}}{x^{1/2}}}}
= ∫ 1 64 x − 1 / 2 + x 1 / 3 − 1 / 2 {\displaystyle \int _{1}^{64}x^{-1/2}+x^{1/3-1/2}} = ∫ 1 64 x − 1 / 2 + x − 1 / 6 {\displaystyle \int _{1}^{64}x^{-1/2}+x^{-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^{1/2}}{\frac {1}{2}}}+{\frac {x^{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}}}
∫ 1 + t a n 2 x ∗ d x = d d x ( t a n ( x ) ) {\displaystyle \int {}{}1+tan^{2}x*dx={\frac {d}{dx}}(tan(x))}
d d x ( t a n ( x ) ) = d d x ( s i n ( x ) c o s ( x ) ) = c o s 2 x − ( − ( s i n ( x ) ) ( s i n ( x ) ) c o s 2 ( x ) {\displaystyle {\frac {d}{dx}}(tan(x))={\frac {d}{dx}}\left({\frac {sin(x)}{cos(x)}}\right)={\frac {cos^{2}x-(-(sin(x))(sin(x))}{cos^{2}(x)}}}
= c o s 2 x + s i n 2 ( x ) ) c o s 2 ( x ) = c o s 2 x c o s 2 ( x ) + s i n 2 x c o s 2 ( x ) {\displaystyle ={\frac {cos^{2}x+sin^{2}(x))}{cos^{2}(x)}}={\frac {cos^{2}x}{cos^{2}(x)}}+{\frac {sin^{2}x}{cos^{2}(x)}}}
= 1 + t a n 2 ( x ) {\displaystyle =1+tan^{2}(x)}
![{\displaystyle \int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cdb02981593f318bab17caae838949d55b200cb6)
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