What is the integral of (e^{-1})/x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{-1})/x

integral

\frac{e ^{- 1}}{x}

\frac{1}{e} ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

= e^{- 1} \cdot \int \frac{1}{x} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{e} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{e} ln ∣ x ∣ + C

(x - 1)^{2}

(e^{x} + e^{- x})^{4}

\frac{1}{( 1 + a ^{2} t ^{2} )}

(3^{x} + 3^{- x})^{2}

(\frac{1 - 1}{n})^{n - 1}