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

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

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

\frac{3}{4} e^{\frac{4 x}{3}} + C

Solution steps

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{3} = 3^{- 1}

= \int ((e^{x})^{3^{- 1}})^{4} d x

Simplify

((e^{x})^{3^{- 1}})^{4} : e^{\frac{4 x}{3}}

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

Apply u-substitution

: \int e^{u} \frac{3}{4} d u

= \int e^{u} \frac{3}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{3}{4} e^{u}

Substitute back

u = \frac{4 x}{3}

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

Add a constant to the solution

= \frac{3}{4} e^{\frac{4 x}{3}} + C

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