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

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

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

\int 3 2 - 5 x^{2} (- 10 x) d x

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

Solution steps

\int 3 2 - 5 x^{2} (- 10 x) d x

Simplify

= \int - 10 x 3 2 - 5 x^{2} d x

Take the constant out:

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

= - 10 \cdot \int x 3 2 - 5 x^{2} d x

Apply u-substitution

= - 10 \cdot \int - \frac{3 u}{10} d u

Take the constant out:

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

= - 10 (- \frac{1}{10} \cdot \int 3 u d u)

Apply the Power Rule

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

Substitute back

u = 2 - 5 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{R} x^{- \frac{9}{5}} d x

\int \frac{4}{x} - \frac{4 x}{x ^{2} + 1} d x

(2 x + 1)^{^{'}}

\int t^{2} sin (t) d t

(x + 2)^{2} y^{^{'}} + 3 (x + 2) y = 4

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