What is the integral of x\sqrt[3]{7-6x^2} ?

The integral of x\sqrt[3]{7-6x^2} is -1/16 (7-6x^2)^{4/3}+C

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integral of x\sqrt[3]{7-6x^2}

\int x 3 7 - 6 x^{2} d x

- \frac{1}{16} (7 - 6 x^{2})^{\frac{4}{3}} + C

Solution steps

\int x 3 7 - 6 x^{2} d x

Apply u-substitution

: \int - \frac{3 u}{12} d u

= \int - \frac{3 u}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int 3 u d u

Apply the Power Rule

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

= - \frac{1}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 7 - 6 x^{2}

= - \frac{1}{12} \cdot \frac{3}{4} (7 - 6 x^{2})^{\frac{4}{3}}

Simplify

- \frac{1}{12} \cdot \frac{3}{4} (7 - 6 x^{2})^{\frac{4}{3}} : - \frac{1}{16} (7 - 6 x^{2})^{\frac{4}{3}}

= - \frac{1}{16} (7 - 6 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{1}{16} (7 - 6 x^{2})^{\frac{4}{3}} + C

\frac{d}{d x} (csc^{2} (2 x) - csc (2) x^{2})

\frac{d}{d x} (e^{\frac{1}{x}} - 1)

\frac{d}{d x} (\frac{5}{x - 2} - 2)

y = cot (2) (sin (x))

\frac{d}{d x} (arccos (2 x))

What is the integral of x\sqrt[3]{7-6x^2} ?
The integral of x\sqrt[3]{7-6x^2} is -1/16 (7-6x^2)^{4/3}+C