What is the integral of (x^{3/2}-x^{1/3})/(6x^{1/4)} ?

The integral of (x^{3/2}-x^{1/3})/(6x^{1/4)} is 2/351 (-27x^{13/12}+13x^{9/4})+C

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integral of (x^{3/2}-x^{1/3})/(6x^{1/4)}

\int \frac{x ^{\frac{3}{2}} - x ^{\frac{1}{3}}}{6 x ^{\frac{1}{4}}} d x

\frac{2}{351} (- 27 x^{\frac{13}{12}} + 13 x^{\frac{9}{4}}) + C

Solution steps

\int \frac{x ^{\frac{3}{2}} - x ^{\frac{1}{3}}}{6 x ^{\frac{1}{4}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{x ^{\frac{3}{2}} - x ^{\frac{1}{3}}}{x ^{\frac{1}{4}}} d x

Expand

\frac{x ^{\frac{3}{2}} - x ^{\frac{1}{3}}}{x ^{\frac{1}{4}}} : x^{\frac{5}{4}} - x^{\frac{1}{12}}

= \frac{1}{6} \cdot \int x^{\frac{5}{4}} - x^{\frac{1}{12}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int x^{\frac{5}{4}} d x - \int x^{\frac{1}{12}} d x)

\int x^{\frac{5}{4}} d x = \frac{4}{9} x^{\frac{9}{4}}

\int x^{\frac{1}{12}} d x = \frac{12}{13} x^{\frac{13}{12}}

= \frac{1}{6} (\frac{4}{9} x^{\frac{9}{4}} - \frac{12}{13} x^{\frac{13}{12}})

Simplify

\frac{1}{6} (\frac{4}{9} x^{\frac{9}{4}} - \frac{12}{13} x^{\frac{13}{12}}) : \frac{2}{351} (- 27 x^{\frac{13}{12}} + 13 x^{\frac{9}{4}})

= \frac{2}{351} (- 27 x^{\frac{13}{12}} + 13 x^{\frac{9}{4}})

Add a constant to the solution

= \frac{2}{351} (- 27 x^{\frac{13}{12}} + 13 x^{\frac{9}{4}}) + C

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What is the integral of (x^{3/2}-x^{1/3})/(6x^{1/4)} ?
The integral of (x^{3/2}-x^{1/3})/(6x^{1/4)} is 2/351 (-27x^{13/12}+13x^{9/4})+C