What is the integral of \sqrt[8]{e^x} ?

The integral of \sqrt[8]{e^x} is 8e^{x/8}+C

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integral of \sqrt[8]{e^x}

8 e^{\frac{x}{8}} + C

Solution steps

Apply radical rule:

assuming

a \geq 0

= \int e^{\frac{x}{8}} d x

Apply u-substitution

: \int e^{u} \cdot 8 d u

= \int e^{u} \cdot 8 d u

Take the constant out:

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

= 8 \cdot \int e^{u} d u

Use the common integral:

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

= 8 e^{u}

Substitute back

u = \frac{x}{8}

= 8 e^{\frac{x}{8}}

Add a constant to the solution

= 8 e^{\frac{x}{8}} + C

x \to 0 lim (\frac{x - y}{x + y})

f (x) = \frac{x ^{2}}{9 + 4 x}

\int \frac{5}{sin ( x )} d x

4 2 x^{2} + 9

f (x) = - 3 x^{5} cos (x)

What is the integral of \sqrt[8]{e^x} ?
The integral of \sqrt[8]{e^x} is 8e^{x/8}+C