What is the integral of (12x)/(\sqrt[3]{3-x^2)} ?

The integral of (12x)/(\sqrt[3]{3-x^2)} is -9(-x^2+3)^{2/3}+C

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integral of (12x)/(\sqrt[3]{3-x^2)}

\int \frac{12 x}{3 3 - x ^{2}} d x

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

Solution steps

\int \frac{12 x}{3 3 - x ^{2}} d x

Take the constant out:

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

= 12 \cdot \int \frac{x}{3 3 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 3 u} d u

= 12 \cdot \int - \frac{1}{2 3 u} d u

Take the constant out:

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

= 12 (- \frac{1}{2} \cdot \int \frac{1}{3 u} d u)

Apply the Power Rule

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

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

Substitute back

u = 3 - x^{2}

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

Simplify

12 (- \frac{1}{2} \cdot \frac{3}{2} (3 - x^{2})^{\frac{2}{3}}) : - 9 (- x^{2} + 3)^{\frac{2}{3}}

= - 9 (- x^{2} + 3)^{\frac{2}{3}}

Add a constant to the solution

= - 9 (- x^{2} + 3)^{\frac{2}{3}} + C

(- 5)^{^{'}}

\int_{5}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

x \to \infty lim (x + 2)

\frac{\partial}{\partial x} (x y + \frac{y}{x})

f (x) = x - 8, at x = 9

What is the integral of (12x)/(\sqrt[3]{3-x^2)} ?
The integral of (12x)/(\sqrt[3]{3-x^2)} is -9(-x^2+3)^{2/3}+C