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

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

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integral of \sqrt[3]{3-4x^2}(-8x)

\int 3 3 - 4 x^{2} (- 8 x) d x

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

Solution steps

\int 3 3 - 4 x^{2} (- 8 x) d x

Simplify

= \int - 8 x 3 3 - 4 x^{2} d x

Take the constant out:

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

= - 8 \cdot \int x 3 3 - 4 x^{2} d x

Apply u-substitution

: \int - \frac{3 u}{8} d u

= - 8 \cdot \int - \frac{3 u}{8} d u

Take the constant out:

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

= - 8 (- \frac{1}{8} \cdot \int 3 u d u)

Apply the Power Rule

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

= - 8 (- \frac{1}{8} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 3 - 4 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int 15 tan^{2} (x se) c^{3} x

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

n = 0 \sum \infty n \cdot sin (\frac{1}{n})

\frac{\partial}{\partial x} (sec^{2} (x y))

n = 1 \sum \infty \frac{3}{π ^{n}}

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