What is the integral of 24x\sqrt[3]{1-3x^2} ?

The integral of 24x\sqrt[3]{1-3x^2} is -3(-3x^2+1)^{4/3}+C

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integral of 24x\sqrt[3]{1-3x^2}

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

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

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

= 24 \cdot \frac{1}{2} (- \frac{1}{3} \cdot \frac{3}{4} v^{\frac{4}{3}})

Substitute back

= 24 \cdot \frac{1}{2} (- \frac{1}{3} \cdot \frac{3}{4} (1 - 3 x^{2})^{\frac{4}{3}})

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 24x\sqrt[3]{1-3x^2} ?
The integral of 24x\sqrt[3]{1-3x^2} is -3(-3x^2+1)^{4/3}+C