What is the integral of x^2\sqrt[5]{7-4x^3} ?

The integral of x^2\sqrt[5]{7-4x^3} is -5/72 (7-4x^3)^{6/5}+C

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integral of x^2\sqrt[5]{7-4x^3}

- \frac{5}{72} (7 - 4 x^{3})^{\frac{6}{5}} + 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{5}{6} u^{\frac{6}{5}}

= - \frac{1}{12} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 7 - 4 x^{3}

= - \frac{1}{12} \cdot \frac{5}{6} (7 - 4 x^{3})^{\frac{6}{5}}

Simplify

- \frac{1}{12} \cdot \frac{5}{6} (7 - 4 x^{3})^{\frac{6}{5}} : - \frac{5}{72} (7 - 4 x^{3})^{\frac{6}{5}}

= - \frac{5}{72} (7 - 4 x^{3})^{\frac{6}{5}}

Add a constant to the solution

= - \frac{5}{72} (7 - 4 x^{3})^{\frac{6}{5}} + C

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What is the integral of x^2\sqrt[5]{7-4x^3} ?
The integral of x^2\sqrt[5]{7-4x^3} is -5/72 (7-4x^3)^{6/5}+C