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

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

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

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

Solution steps

Apply radical rule:

assuming

a \geq 0

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

Apply u-substitution

: \int e^{u} \cdot 4 d u

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

Take the constant out:

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

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

Use the common integral:

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

= 4 e^{u}

Substitute back

u = \frac{x}{4}

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

Add a constant to the solution

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

\int_{- \infty}^{0} \frac{1}{2 - 9 x} d x

\frac{d}{d x} (3 x^{2} + 9 x + 2)

\frac{d y}{d x} = 0

\int \frac{e ^{x + 3}}{x + 3} d x

y = - \frac{1}{2} x^{2} + 2, y = 0

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