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

The integral of (\sqrt[3]{x^4})/4 is 3/28 x^{7/3}+C

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

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

\frac{3}{28} x^{\frac{7}{3}} + C

Solution steps

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

Take the constant out:

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

= \frac{1}{4} \cdot \int 3 x^{4} d x

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

assuming

x \geq 0

= \frac{1}{4} \cdot \int x^{\frac{4}{3}} d x

Apply the Power Rule

: \frac{3}{7} x^{\frac{7}{3}}

= \frac{1}{4} \cdot \frac{3}{7} x^{\frac{7}{3}}

Simplify

\frac{1}{4} \cdot \frac{3}{7} x^{\frac{7}{3}} : \frac{3}{28} x^{\frac{7}{3}}

= \frac{3}{28} x^{\frac{7}{3}}

Add a constant to the solution

= \frac{3}{28} x^{\frac{7}{3}} + C

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What is the integral of (\sqrt[3]{x^4})/4 ?
The integral of (\sqrt[3]{x^4})/4 is 3/28 x^{7/3}+C