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

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

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

\frac{3}{4} (3 - 2 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}}

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

Substitute back

u = 3 - 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 4 y = e^{- 2 x} - 2 x

\int cos (10 x) cos (3 x) d x

f (x) = x + \frac{2}{x}, at x = 3

1500 e^{x}

\frac{d}{d x} (\frac{x ^{2}}{x + 1})

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