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

Take the constant out:

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

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

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 = 1 + e^{x}

= - ln ∣ 1 + e^{x} ∣

Add a constant to the solution

= - ln ∣ 1 + e^{x} ∣ + C

\int \frac{x + 34}{( x - 6 ) ( x + 2 )} d x

\int sin (π w) cos (π w) d w

\int \frac{9 x ^{2}}{1 - x ^{3}} d x

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

\int \frac{2 sin ( x )}{cos ^{3} ( x )} d x

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