What is the integral of (\sqrt[3]{x})^4 ?

The integral of (\sqrt[3]{x})^4 is 1/2 x^2\sqrt[3]{x}-1/14 x^{7/3}+C

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integral of (\sqrt[3]{x})^4

Solution steps

Simplify

Apply Integration By Parts

\int \frac{1}{6} x^{\frac{4}{3}} d x = \frac{1}{14} x^{\frac{7}{3}}

Add a constant to the solution

\frac{d}{d x} (x^{2} lo g_{2} (x^{3}))

u^{5}

\frac{d}{d x} (\frac{x ^{4}}{3} - 7 x)

e^{- 2 t}

\frac{d}{d x} (x 5)

What is the integral of (\sqrt[3]{x})^4 ?
The integral of (\sqrt[3]{x})^4 is 1/2 x^2\sqrt[3]{x}-1/14 x^{7/3}+C