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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{9 + x f ( x )}{x}

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

\frac{d}{d x} (\frac{x}{x ^{2} + 2})

\frac{1}{x} sin (x)

f (x) = x^{4} (3 - x)^{3}, at x = 1

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