What is the integral of 24x(6-x^2)^{1/3} ?

The integral of 24x(6-x^2)^{1/3} is -9(-x^2+6)^{4/3}+C

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integral of 24x(6-x^2)^{1/3}

\int 24 x (6 - x^{2})^{\frac{1}{3}} d x

- 9 (- x^{2} + 6)^{\frac{4}{3}} + C

Solution steps

\int 24 x (6 - x^{2})^{\frac{1}{3}} d x

Take the constant out:

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

= 24 \cdot \int x (6 - x^{2})^{\frac{1}{3}} d x

Apply u-substitution

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

= 24 \cdot \int - \frac{u ^{\frac{1}{3}}}{2} d u

Take the constant out:

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

= 24 (- \frac{1}{2} \cdot \int u^{\frac{1}{3}} d u)

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= 24 (- \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 6 - x^{2}

= 24 (- \frac{1}{2} \cdot \frac{3}{4} (6 - x^{2})^{\frac{4}{3}})

Simplify

24 (- \frac{1}{2} \cdot \frac{3}{4} (6 - x^{2})^{\frac{4}{3}}) : - 9 (- x^{2} + 6)^{\frac{4}{3}}

= - 9 (- x^{2} + 6)^{\frac{4}{3}}

Add a constant to the solution

= - 9 (- x^{2} + 6)^{\frac{4}{3}} + C

f (x) = (x + 1) e^{x}

x^{2} + 1, 5

x^{2} + 11

\frac{c}{x} + \frac{2 a}{x ^{3}} + 8

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

What is the integral of 24x(6-x^2)^{1/3} ?
The integral of 24x(6-x^2)^{1/3} is -9(-x^2+6)^{4/3}+C