What is the integral of (-6x^2)/(\sqrt[3]{3-4x^3)} ?

The integral of (-6x^2)/(\sqrt[3]{3-4x^3)} is 3/4 (-4x^3+3)^{2/3}+C

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integral of (-6x^2)/(\sqrt[3]{3-4x^3)}

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

Solution steps

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}{2} u^{\frac{2}{3}}

= - 6 (- \frac{1}{12} \cdot \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 3 - 4 x^{3}

= - 6 (- \frac{1}{12} \cdot \frac{3}{2} (3 - 4 x^{3})^{\frac{2}{3}})

Simplify

- 6 (- \frac{1}{12} \cdot \frac{3}{2} (3 - 4 x^{3})^{\frac{2}{3}}) : \frac{3}{4} (- 4 x^{3} + 3)^{\frac{2}{3}}

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

Add a constant to the solution

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

\int \frac{x ^{2} + 1}{x ( x ^{2} + 3 )} d x

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

\frac{s + 5}{7 s ^{2} + 26 s + 25}

\frac{d y}{d t} + 4 y = 0

f (x) = x^{7}, at x = 2

What is the integral of (-6x^2)/(\sqrt[3]{3-4x^3)} ?
The integral of (-6x^2)/(\sqrt[3]{3-4x^3)} is 3/4 (-4x^3+3)^{2/3}+C