What is the integral of x^3\sqrt[3]{8-3x^4} ?

The integral of x^3\sqrt[3]{8-3x^4} is -1/16 (8-3x^4)^{4/3}+C

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integral of x^3\sqrt[3]{8-3x^4}

- \frac{1}{16} (8 - 3 x^{4})^{\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

Apply the Power Rule

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

= - \frac{1}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 8 - 3 x^{4}

= - \frac{1}{12} \cdot \frac{3}{4} (8 - 3 x^{4})^{\frac{4}{3}}

Simplify

- \frac{1}{12} \cdot \frac{3}{4} (8 - 3 x^{4})^{\frac{4}{3}} : - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}}

= - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}} + C

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What is the integral of x^3\sqrt[3]{8-3x^4} ?
The integral of x^3\sqrt[3]{8-3x^4} is -1/16 (8-3x^4)^{4/3}+C