What is the integral of x^2\sqrt[7]{x^3-16} ?

The integral of x^2\sqrt[7]{x^3-16} is 7/24 (x^3-16)^{8/7}+C

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integral of x^2\sqrt[7]{x^3-16}

\frac{7}{24} (x^{3} - 16)^{\frac{8}{7}} + 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{7}{8} u^{\frac{8}{7}}

= \frac{1}{3} \cdot \frac{7}{8} u^{\frac{8}{7}}

Substitute back

u = x^{3} - 16

= \frac{1}{3} \cdot \frac{7}{8} (x^{3} - 16)^{\frac{8}{7}}

Simplify

\frac{1}{3} \cdot \frac{7}{8} (x^{3} - 16)^{\frac{8}{7}} : \frac{7}{24} (x^{3} - 16)^{\frac{8}{7}}

= \frac{7}{24} (x^{3} - 16)^{\frac{8}{7}}

Add a constant to the solution

= \frac{7}{24} (x^{3} - 16)^{\frac{8}{7}} + C

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What is the integral of x^2\sqrt[7]{x^3-16} ?
The integral of x^2\sqrt[7]{x^3-16} is 7/24 (x^3-16)^{8/7}+C