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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int n e^{x} d x

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

Solution steps

\int n e^{x} d x

Apply radical rule:

n a^{m} = a^{\frac{m}{n}},

assuming

a \geq 0

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

Apply u-substitution

: \int e^{u} n d u

= \int e^{u} n d u

Take the constant out:

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

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

Use the common integral:

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

= n e^{u}

Substitute back

u = \frac{x}{n}

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

Add a constant to the solution

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

\int (- \frac{2}{x} + 3 cos (6 x)) d x

\int 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

\int \frac{14}{x + 1} d x

\int (x y^{2} + y - x) d x

\int \frac{ln ^{9} ( x )}{x} d x

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