What is the integral of \sqrt[3]{4-3x^2}(-6x) ?

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

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integral of \sqrt[3]{4-3x^2}(-6x)

\int 3 4 - 3 x^{2} (- 6 x) d x

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

Solution steps

\int 3 4 - 3 x^{2} (- 6 x) d x

Simplify

= \int - 6 x 3 4 - 3 x^{2} d x

Take the constant out:

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

= - 6 \cdot \int x 3 4 - 3 x^{2} d x

Apply u-substitution

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

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

Take the constant out:

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

= - 6 (- \frac{1}{6} \cdot \int 3 u d u)

Apply the Power Rule

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

= - 6 (- \frac{1}{6} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 4 - 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} - 2 x - 3)

x \to 2 lim (2 x^{2} + 1)

- \frac{x + 3 arcsin ( x )}{4}

\int \frac{4 x - 10}{x ^{2} - 6 x + 8} d x

\int \frac{x + 2}{x + 5} d x

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