What is the integral of 1/(sqrt(e^x)) ?

The integral of 1/(sqrt(e^x)) is -2/(e^{x/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(e^x))

\int \frac{1}{e ^{x}} d x

- \frac{2}{e ^{\frac{x}{2}}} + C

Solution steps

\int \frac{1}{e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u u} d u

= \int \frac{1}{u u} d u

Simplify

u u : u^{\frac{3}{2}}

= \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u}

= - \frac{2}{u}

Substitute back

u = e^{x}

= - \frac{2}{e ^{x}}

Apply radical rule:

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

assuming

a \geq 0

= - \frac{2}{e ^{\frac{x}{2}}}

Add a constant to the solution

= - \frac{2}{e ^{\frac{x}{2}}} + C

\frac{d x}{d t} = 2 x

x \to \infty lim 1 + \frac{1}{x} - 1 x

y^{^{'}} = e^{x - y}

\int_{- 2}^{2} 5 - 4 - x^{2} d x

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

What is the integral of 1/(sqrt(e^x)) ?
The integral of 1/(sqrt(e^x)) is -2/(e^{x/2)}+C