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

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

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

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

\frac{1}{7} x^{\frac{7}{4}} + C

Solution steps

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

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

assuming

x \geq 0

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

Apply the Power Rule

: \frac{4}{7} x^{\frac{7}{4}}

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{7} x^{\frac{7}{4}} + C

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