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

The integral of (x^3)/(\sqrt[3]{2-x^4)} is -3/8 (2-x^4)^{2/3}+C

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integral of (x^3)/(\sqrt[3]{2-x^4)}

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

Substitute back

u = 2 - x^{4}

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

Simplify

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

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

Add a constant to the solution

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

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