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

The integral of (x^{1/3}-x^{2/3})^2 is -x^2+3/5 x^{5/3}+3/7 x^{7/3}+C

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

\int (x^{\frac{1}{3}} - x^{\frac{2}{3}})^{2} d x

- x^{2} + \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{7} x^{\frac{7}{3}} + C

Solution steps

\int (x^{\frac{1}{3}} - x^{\frac{2}{3}})^{2} d x

= \int (((x^{\frac{1}{3}})^{\frac{1}{2}} + (x^{\frac{2}{3}})^{\frac{1}{2}}) ((x^{\frac{1}{3}})^{\frac{1}{2}} - (x^{\frac{2}{3}})^{\frac{1}{2}}))^{2} d x

Simplify

(((x^{\frac{1}{3}})^{\frac{1}{2}} + (x^{\frac{2}{3}})^{\frac{1}{2}}) ((x^{\frac{1}{3}})^{\frac{1}{2}} - (x^{\frac{2}{3}})^{\frac{1}{2}}))^{2} : - 2 x + x^{\frac{2}{3}} + x^{\frac{4}{3}}

= \int - 2 x + x^{\frac{2}{3}} + x^{\frac{4}{3}} 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

= - \int 2 x d x + \int x^{\frac{2}{3}} d x + \int x^{\frac{4}{3}} d x

\int 2 x d x = x^{2}

\int x^{\frac{2}{3}} d x = \frac{3}{5} x^{\frac{5}{3}}

\int x^{\frac{4}{3}} d x = \frac{3}{7} x^{\frac{7}{3}}

= - x^{2} + \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{7} x^{\frac{7}{3}}

Add a constant to the solution

= - x^{2} + \frac{3}{5} x^{\frac{5}{3}} + \frac{3}{7} x^{\frac{7}{3}} + C

csc (x) cot (x)

y = (sin (3 x))^{l n (x)}

\frac{\partial}{\partial y} (ln (y^{x}))

t \to - 5 lim (t^{2} - 5)

y = 6 x sin (x), (\frac{π}{2}, 3 π)

What is the integral of (x^{1/3}-x^{2/3})^2 ?
The integral of (x^{1/3}-x^{2/3})^2 is -x^2+3/5 x^{5/3}+3/7 x^{7/3}+C