What is the integral of (8x(8-3x^2)^{1/3})/3 ?

The integral of (8x(8-3x^2)^{1/3})/3 is -1/3 (8-3x^2)^{4/3}+C

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integral of (8x(8-3x^2)^{1/3})/3

\int \frac{8 x ( 8 - 3 x ^{2} ) ^{\frac{1}{3}}}{3} d x

- \frac{1}{3} (8 - 3 x^{2})^{\frac{4}{3}} + C

Solution steps

\int \frac{8 x ( 8 - 3 x ^{2} ) ^{\frac{1}{3}}}{3} d x

Take the constant out:

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

= \frac{8}{3} \cdot \int x (8 - 3 x^{2})^{\frac{1}{3}} d x

Apply u-substitution

: \int - \frac{u ^{\frac{1}{3}}}{6} d u

= \frac{8}{3} \cdot \int - \frac{u ^{\frac{1}{3}}}{6} d u

Take the constant out:

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

= \frac{8}{3} (- \frac{1}{6} \cdot \int u^{\frac{1}{3}} d u)

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{8}{3} (- \frac{1}{6} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 8 - 3 x^{2}

= \frac{8}{3} (- \frac{1}{6} \cdot \frac{3}{4} (8 - 3 x^{2})^{\frac{4}{3}})

Simplify

\frac{8}{3} (- \frac{1}{6} \cdot \frac{3}{4} (8 - 3 x^{2})^{\frac{4}{3}}) : - \frac{1}{3} (8 - 3 x^{2})^{\frac{4}{3}}

= - \frac{1}{3} (8 - 3 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{1}{3} (8 - 3 x^{2})^{\frac{4}{3}} + C

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What is the integral of (8x(8-3x^2)^{1/3})/3 ?
The integral of (8x(8-3x^2)^{1/3})/3 is -1/3 (8-3x^2)^{4/3}+C