What is the integral of x\sqrt[3]{x^2-9} ?

The integral of x\sqrt[3]{x^2-9} is 3/8 (x^2-9)^{4/3}+C

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integral of x\sqrt[3]{x^2-9}

\frac{3}{8} (x^{2} - 9)^{\frac{4}{3}} + C

Solution steps

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}}

= \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x^{2} - 9

= \frac{1}{2} \cdot \frac{3}{4} (x^{2} - 9)^{\frac{4}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{4} (x^{2} - 9)^{\frac{4}{3}} : \frac{3}{8} (x^{2} - 9)^{\frac{4}{3}}

= \frac{3}{8} (x^{2} - 9)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{8} (x^{2} - 9)^{\frac{4}{3}} + C

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

x \to 4 - lim (\frac{x - 10}{x - 4})

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

\frac{\partial}{\partial y} (\frac{1}{2 - x + y})

x^{2} + 3 y \cdot \frac{d y}{d x} = 0

What is the integral of x\sqrt[3]{x^2-9} ?
The integral of x\sqrt[3]{x^2-9} is 3/8 (x^2-9)^{4/3}+C