What is the integral of x^2(\sqrt[4]{x^3-1}) ?

The integral of x^2(\sqrt[4]{x^3-1}) is 4/15 (x^3-1)^{5/4}+C

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integral of x^2(\sqrt[4]{x^3-1})

\frac{4}{15} (x^{3} - 1)^{\frac{5}{4}} + 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{4}{5} u^{\frac{5}{4}}

= \frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = x^{3} - 1

= \frac{1}{3} \cdot \frac{4}{5} (x^{3} - 1)^{\frac{5}{4}}

Simplify

\frac{1}{3} \cdot \frac{4}{5} (x^{3} - 1)^{\frac{5}{4}} : \frac{4}{15} (x^{3} - 1)^{\frac{5}{4}}

= \frac{4}{15} (x^{3} - 1)^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{15} (x^{3} - 1)^{\frac{5}{4}} + C

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What is the integral of x^2(\sqrt[4]{x^3-1}) ?
The integral of x^2(\sqrt[4]{x^3-1}) is 4/15 (x^3-1)^{5/4}+C