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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(e^x+1)

integral

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

- ln ∣ e^{x} + 1 ∣ + x + C

Solution steps

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{e ^{x}}{e ^{x} + 1} d x = ln ∣ e^{x} + 1 ∣

\int 1 d x = x

= - ln ∣ e^{x} + 1 ∣ + x

Add a constant to the solution

= - ln ∣ e^{x} + 1 ∣ + x + C

\frac{( 2 x + 5 )}{( x + 2 )}

sin (2 x + 1)

x + 16

\frac{1}{( 35 - x )}

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