What is the integral of \sqrt[6]{e^x} ?

The integral of \sqrt[6]{e^x} is 6e^{x/6}+C

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integral of \sqrt[6]{e^x}

6 e^{\frac{x}{6}} + C

Solution steps

Apply radical rule:

assuming

a \geq 0

= \int e^{\frac{x}{6}} d x

Apply u-substitution

: \int e^{u} \cdot 6 d u

= \int e^{u} \cdot 6 d u

Take the constant out:

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

= 6 \cdot \int e^{u} d u

Use the common integral:

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

= 6 e^{u}

Substitute back

u = \frac{x}{6}

= 6 e^{\frac{x}{6}}

Add a constant to the solution

= 6 e^{\frac{x}{6}} + C

f (x) = \frac{1}{x + 1}

f (x) = x^{2} + 4, (- 4, 20)

\frac{d}{d u} (2 uv)

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

y^{^{'}} + e^{y} t = e^{y} sin (t)

What is the integral of \sqrt[6]{e^x} ?
The integral of \sqrt[6]{e^x} is 6e^{x/6}+C