What is the integral of \sqrt[3]{2-3x^2}(-6x) ?

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

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integral of \sqrt[3]{2-3x^2}(-6x)

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

Solution steps

Simplify

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

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

Substitute back

u = 2 - 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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