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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 3 - 2 x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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