What is the integral of x^2*\sqrt[3]{1-x} ?

The integral of x^2*\sqrt[3]{1-x} is -3/10 (1-x)^{10/3}+6/7 (1-x)^{7/3}-3/4 (1-x)^{4/3}+C

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integral of x^2*\sqrt[3]{1-x}

- \frac{3}{10} (1 - x)^{\frac{10}{3}} + \frac{6}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

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

\int u^{\frac{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

\int 2 u^{\frac{4}{3}} d u = \frac{6}{7} u^{\frac{7}{3}}

= - (\frac{3}{10} u^{\frac{10}{3}} - \frac{6}{7} u^{\frac{7}{3}} + \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 1 - x

= - (\frac{3}{10} (1 - x)^{\frac{10}{3}} - \frac{6}{7} (1 - x)^{\frac{7}{3}} + \frac{3}{4} (1 - x)^{\frac{4}{3}})

Simplify

= - \frac{3}{10} (1 - x)^{\frac{10}{3}} + \frac{6}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}}

Add a constant to the solution

= - \frac{3}{10} (1 - x)^{\frac{10}{3}} + \frac{6}{7} (1 - x)^{\frac{7}{3}} - \frac{3}{4} (1 - x)^{\frac{4}{3}} + C

What is the integral of x^2*\sqrt[3]{1-x} ?
The integral of x^2*\sqrt[3]{1-x} is -3/10 (1-x)^{10/3}+6/7 (1-x)^{7/3}-3/4 (1-x)^{4/3}+C