What is the integral of \sqrt[9]{e^x} ?

The integral of \sqrt[9]{e^x} is 9e^{x/9}+C

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integral of \sqrt[9]{e^x}

9 e^{\frac{x}{9}} + C

Solution steps

Apply radical rule:

assuming

a \geq 0

= \int e^{\frac{x}{9}} d x

Apply u-substitution

: \int e^{u} \cdot 9 d u

= \int e^{u} \cdot 9 d u

Take the constant out:

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

= 9 \cdot \int e^{u} d u

Use the common integral:

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

= 9 e^{u}

Substitute back

u = \frac{x}{9}

= 9 e^{\frac{x}{9}}

Add a constant to the solution

= 9 e^{\frac{x}{9}} + C

(y^{2} + 1) d x = y sec^{2} (x) d y

x \to 0 lim (x^{- 2})

(2 x) d x + (2 y - 4 x^{2} y^{- 1}) d y = 0

\frac{d}{d x} (ln x^{2} - 1)

\int \frac{u}{( u ^{2} + 1 ) ( 1 - 2 u ) ( 1 + 2 u )} d u

What is the integral of \sqrt[9]{e^x} ?
The integral of \sqrt[9]{e^x} is 9e^{x/9}+C