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

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

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

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

d er i v a t i v e y = cot (2) (sin (x))

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

n = 1 \sum \infty \frac{1}{( 3 n + 5 ) ^{4}}

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