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

The integral of \sqrt[3]{6-2x} is -3/8 (6-2x)^{4/3}+C

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

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

Substitute back

u = 6 - 2 x

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

Simplify

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

= - \frac{3}{8} (6 - 2 x)^{\frac{4}{3}}

Add a constant to the solution

= - \frac{3}{8} (6 - 2 x)^{\frac{4}{3}} + C

What is the integral of \sqrt[3]{6-2x} ?
The integral of \sqrt[3]{6-2x} is -3/8 (6-2x)^{4/3}+C