What is the integral of \sqrt[3]{4-5x^2}(-10x) ?

The integral of \sqrt[3]{4-5x^2}(-10x) is 3/4 (4-5x^2)^{4/3}+C

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integral of \sqrt[3]{4-5x^2}(-10x)

\frac{3}{4} (4 - 5 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}}

= - 10 (- \frac{1}{10} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 4 - 5 x^{2}

= - 10 (- \frac{1}{10} \cdot \frac{3}{4} (4 - 5 x^{2})^{\frac{4}{3}})

Simplify

- 10 (- \frac{1}{10} \cdot \frac{3}{4} (4 - 5 x^{2})^{\frac{4}{3}}) : \frac{3}{4} (4 - 5 x^{2})^{\frac{4}{3}}

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

Add a constant to the solution

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

n = 1 \sum \infty 9 e^{- n}

\int_{- 1}^{3} x e^{x y} d y

- x + 4, 2 x, x

\frac{d}{d x} (e^{t x})

\int_{4}^{9} 4 x d x

What is the integral of \sqrt[3]{4-5x^2}(-10x) ?
The integral of \sqrt[3]{4-5x^2}(-10x) is 3/4 (4-5x^2)^{4/3}+C