What is the integral of (6\sqrt[3]{x}) ?

The integral of (6\sqrt[3]{x}) is 9/2 x^{4/3}+C

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integral of (6\sqrt[3]{x})

\frac{9}{2} x^{\frac{4}{3}} + C

Solution steps

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} x^{\frac{4}{3}}

= 6 \cdot \frac{3}{4} x^{\frac{4}{3}}

Simplify

6 \cdot \frac{3}{4} x^{\frac{4}{3}} : \frac{9}{2} x^{\frac{4}{3}}

= \frac{9}{2} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{9}{2} x^{\frac{4}{3}} + C

\frac{d}{d x} (20 x (2 x^{2} + 7)^{4})

2 + 6 x^{2} - 2 x^{3}

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

\frac{\partial}{\partial x} (a x^{2} + b y^{2})

\frac{d}{d x} (\frac{x}{x ^{2} + 81})

What is the integral of (6\sqrt[3]{x}) ?
The integral of (6\sqrt[3]{x}) is 9/2 x^{4/3}+C