What is the integral of x\sqrt[3]{16-3x} ?

The integral of x\sqrt[3]{16-3x} is 1/21 ((-3x+16)^{7/3}-28(-3x+16)^{4/3})+C

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integral of x\sqrt[3]{16-3x}

\frac{1}{21} ((- 3 x + 16)^{\frac{7}{3}} - 28 (- 3 x + 16)^{\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{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= \frac{1}{9} (\frac{3}{7} u^{\frac{7}{3}} - 12 u^{\frac{4}{3}})

Substitute back

u = 16 - 3 x

= \frac{1}{9} (\frac{3}{7} (16 - 3 x)^{\frac{7}{3}} - 12 (16 - 3 x)^{\frac{4}{3}})

Simplify

\frac{1}{9} (\frac{3}{7} (16 - 3 x)^{\frac{7}{3}} - 12 (16 - 3 x)^{\frac{4}{3}}) : \frac{1}{21} ((- 3 x + 16)^{\frac{7}{3}} - 28 (- 3 x + 16)^{\frac{4}{3}})

= \frac{1}{21} ((- 3 x + 16)^{\frac{7}{3}} - 28 (- 3 x + 16)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{21} ((- 3 x + 16)^{\frac{7}{3}} - 28 (- 3 x + 16)^{\frac{4}{3}}) + C

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What is the integral of x\sqrt[3]{16-3x} ?
The integral of x\sqrt[3]{16-3x} is 1/21 ((-3x+16)^{7/3}-28(-3x+16)^{4/3})+C