What is the integral of \sqrt[5]{e^x} ?

The integral of \sqrt[5]{e^x} is 5e^{x/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[5]{e^x}

5 e^{\frac{x}{5}} + C

Solution steps

Apply radical rule:

assuming

a \geq 0

= \int e^{\frac{x}{5}} d x

Apply u-substitution

: \int e^{u} \cdot 5 d u

= \int e^{u} \cdot 5 d u

Take the constant out:

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

= 5 \cdot \int e^{u} d u

Use the common integral:

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

= 5 e^{u}

Substitute back

u = \frac{x}{5}

= 5 e^{\frac{x}{5}}

Add a constant to the solution

= 5 e^{\frac{x}{5}} + C

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

5 x + 11

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

x \to 0 lim (\frac{( x + 64 ) ^{\frac{1}{3}} - 4}{x})

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

What is the integral of \sqrt[5]{e^x} ?
The integral of \sqrt[5]{e^x} is 5e^{x/5}+C