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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^y)/(e^y+1)

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

ln ∣ e^{y} + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = e^{y} + 1

= ln ∣ e^{y} + 1 ∣

Add a constant to the solution

= ln ∣ e^{y} + 1 ∣ + C

\int \frac{1}{5 x ^{3} - 7 x ^{2}} d x

x^{2}, x^{k} 0, 1

\int_{- \infty}^{0} \frac{1}{7 - 3 x} d x

\int \frac{20}{( 100 x ^{2} + 1 ) ^{2}} d x

\frac{d}{d t} (sin (t) - t cos (t))

What is the integral of (e^y)/(e^y+1) ?
The integral of (e^y)/(e^y+1) is ln|e^y+1|+C