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

The integral of x\sqrt[3]{3-x^2} is -3/8 (3-x^2)^{4/3}+C

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

- \frac{3}{8} (3 - x^{2})^{\frac{4}{3}} + 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{3}{4} u^{\frac{4}{3}}

= - \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 3 - x^{2}

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

Simplify

- \frac{1}{2} \cdot \frac{3}{4} (3 - x^{2})^{\frac{4}{3}} : - \frac{3}{8} (3 - x^{2})^{\frac{4}{3}}

= - \frac{3}{8} (3 - x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{3}{8} (3 - x^{2})^{\frac{4}{3}} + C

f (x) = 2 - 4 x

4, \frac{e ^{x}}{5}, x = 0

\frac{5}{x}, at x = \frac{1}{4}

cos (x - 1)

What is the integral of x\sqrt[3]{3-x^2} ?
The integral of x\sqrt[3]{3-x^2} is -3/8 (3-x^2)^{4/3}+C