What is the integral of \sqrt[3]{5-4x^2}(-8x) ?

The integral of \sqrt[3]{5-4x^2}(-8x) is 3/4 (5-4x^2)^{4/3}+C

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integral of \sqrt[3]{5-4x^2}(-8x)

\frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}} + C

Solution steps

Simplify

Take the constant out:

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

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}}

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

Substitute back

u = 5 - 4 x^{2}

= - 8 (- \frac{1}{8} \cdot \frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}})

Simplify

- 8 (- \frac{1}{8} \cdot \frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}}) : \frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}}

= \frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} (5 - 4 x^{2})^{\frac{4}{3}} + C

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\frac{\partial}{\partial x} (arctan (2 x y))

\int \frac{( e ^{x} - e ^{- x} )}{e ^{2 x}} d x

\frac{\partial}{\partial s} (s cos (t))

\frac{d}{d x} (x^{2} - 4 - 2 arcsec (\frac{x}{2}))

What is the integral of \sqrt[3]{5-4x^2}(-8x) ?
The integral of \sqrt[3]{5-4x^2}(-8x) is 3/4 (5-4x^2)^{4/3}+C