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

\int [3 e^{x}]^{4} d x

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

Solution steps

\int [3 e^{x}]^{4} d x

Simplify

[3 e^{x}]^{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

\frac{\partial}{\partial x} (x^{2} e^{5 x y})

x \to 0 - lim (csc (x) - cot (x))

\int \frac{6}{z ^{4}} d z

x \to 4 lim (\frac{x}{x})

f (x) = 4 x^{2} - 2 x, at x = - 1

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