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

The integral of (\sqrt[3]{3x})/2 is 1/2 (\sqrt[3]{3}x\sqrt[3]{x}-3/(4\sqrt[3]{9)}x^{4/3})+C

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

\int \frac{3 3 x}{2} d x

\frac{1}{2} (33 x 3 x - \frac{3}{4 3 9} x^{\frac{4}{3}}) + C

Solution steps

\int \frac{3 3 x}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int 3 3 x d x

Apply Integration By Parts

= \frac{1}{2} (33 x 3 x - \int 3 \frac{x}{9} d x)

\int 3 \frac{x}{9} d x = \frac{3}{4 3 9} x^{\frac{4}{3}}

= \frac{1}{2} (33 x 3 x - \frac{3}{4 3 9} x^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{2} (33 x 3 x - \frac{3}{4 3 9} x^{\frac{4}{3}}) + C

What is the integral of (\sqrt[3]{3x})/2 ?
The integral of (\sqrt[3]{3x})/2 is 1/2 (\sqrt[3]{3}x\sqrt[3]{x}-3/(4\sqrt[3]{9)}x^{4/3})+C