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

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

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

\int (6 x)^{3} d x

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

Solution steps

\int (6 x)^{3} d x

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

(6 x)^{3} = x

= \int x d x

Apply the Power Rule

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

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

Add a constant to the solution

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

n = 1 \sum \infty 4^{n} \cdot x^{n}

n = 1 \sum \infty \frac{1}{3 n + 1}

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

\int sin (- 5 x^{4} - x - 6) (- 20 x^{3} - 1) d x

\int \frac{x}{1 - 3 x ^{2}} d x

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