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

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

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

- ln ∣ e^{y} + 1 ∣ + y + C

Solution steps

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

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

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

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 ^{y}}{e ^{y} + 1} d y + \int 1 d y

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

\int 1 d y = y

= - ln ∣ e^{y} + 1 ∣ + y

Add a constant to the solution

= - ln ∣ e^{y} + 1 ∣ + y + C

\int 3 sin (3 s) d s

\int \frac{sec ^{3} ( θ )}{tan ^{4} ( θ )} d θ

\int \frac{1}{x ^{2} + \frac{2}{3} x + \frac{2}{3}} d x

\int (x^{3} (x^{4} + 3)^{2}) d x

\int x^{5} (x^{6} - 6)^{4} d x