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

The integral of \sqrt[3]{3-5x^2}x is -3/40 (3-5x^2)^{4/3}+C

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

\int 3 3 - 5 x^{2} x d x

- \frac{3}{40} (3 - 5 x^{2})^{\frac{4}{3}} + C

Solution steps

\int 3 3 - 5 x^{2} x d x

Apply u-substitution

: \int - \frac{3 u}{10} d u

= \int - \frac{3 u}{10} d u

Take the constant out:

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

= - \frac{1}{10} \cdot \int 3 u d u

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= - \frac{1}{10} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 3 - 5 x^{2}

= - \frac{1}{10} \cdot \frac{3}{4} (3 - 5 x^{2})^{\frac{4}{3}}

Simplify

- \frac{1}{10} \cdot \frac{3}{4} (3 - 5 x^{2})^{\frac{4}{3}} : - \frac{3}{40} (3 - 5 x^{2})^{\frac{4}{3}}

= - \frac{3}{40} (3 - 5 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{3}{40} (3 - 5 x^{2})^{\frac{4}{3}} + C

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What is the integral of \sqrt[3]{3-5x^2}x ?
The integral of \sqrt[3]{3-5x^2}x is -3/40 (3-5x^2)^{4/3}+C