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

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

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

\int \frac{6}{3 x} d x

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

Solution steps

\int \frac{6}{3 x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{3 x} d x

Apply the Power Rule

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

= 6 \cdot \frac{3}{2} x^{\frac{2}{3}}

Simplify

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

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

Add a constant to the solution

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

y = t^{3} - 2 t + 4

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 5

x \to 0 lim (\frac{( 2 + x ) ^{2} + 4}{x})

\frac{1}{x ( 5 + ln ( x ) )}

\frac{d y}{d x} = 2 e^{(x - y)}

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