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

The integral of \sqrt[3]{(7-2x^3)^4}x^2 is -1/14 (7-2x^3)^{7/3}+C

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

\int 3 (7 - 2 x^{3})^{4} x^{2} d x

- \frac{1}{14} (7 - 2 x^{3})^{\frac{7}{3}} + C

Solution steps

\int 3 (7 - 2 x^{3})^{4} x^{2} d x

Apply u-substitution

: \int - \frac{u ^{\frac{4}{3}}}{6} d u

= \int - \frac{u ^{\frac{4}{3}}}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{6} \cdot \int u^{\frac{4}{3}} d u

Apply the Power Rule

: \frac{3}{7} u^{\frac{7}{3}}

= - \frac{1}{6} \cdot \frac{3}{7} u^{\frac{7}{3}}

Substitute back

u = 7 - 2 x^{3}

= - \frac{1}{6} \cdot \frac{3}{7} (7 - 2 x^{3})^{\frac{7}{3}}

Simplify

- \frac{1}{6} \cdot \frac{3}{7} (7 - 2 x^{3})^{\frac{7}{3}} : - \frac{1}{14} (7 - 2 x^{3})^{\frac{7}{3}}

= - \frac{1}{14} (7 - 2 x^{3})^{\frac{7}{3}}

Add a constant to the solution

= - \frac{1}{14} (7 - 2 x^{3})^{\frac{7}{3}} + C

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What is the integral of \sqrt[3]{(7-2x^3)^4}x^2 ?
The integral of \sqrt[3]{(7-2x^3)^4}x^2 is -1/14 (7-2x^3)^{7/3}+C