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

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

\int \frac{( e ^{- x} )}{1 + e ^{- x}} d x

- ln 1 + e^{- x} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 + e^{- x}

= - ln 1 + e^{- x}

Add a constant to the solution

= - ln 1 + e^{- x} + C

\int 2 b e^{2 a x} d x

\int \frac{3 x + 2}{x ^{2} + x + 2} d x

\int 3 x csc (3 x) cot (3 x) d x

\int \frac{- 9 x - 9}{x ( x ^{2} - 9 )} d x

\int x^{4} e^{x y} d y

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