What is the integral of (e^{-x})/(1+e^{2x)} ?

The integral of (e^{-x})/(1+e^{2x)} is arctan(e^{-x})-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^{2x)}

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

arctan (e^{- x}) - e^{- x} + C

Solution steps

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

Apply u-substitution

: \int - \frac{e ^{u}}{1 + e ^{- 2 u}} d u

= \int - \frac{e ^{u}}{1 + e ^{- 2 u}} d u

Take the constant out:

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

= - \int \frac{e ^{u}}{1 + e ^{- 2 u}} d u

Apply u-substitution

: \int \frac{v ^{2}}{v ^{2} + 1} d v

= - \int \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= - \int - \frac{1}{v ^{2} + 1} + 1 d v

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{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= - (- arctan (v) + v)

Substitute back

= - (- arctan (e^{- x}) + e^{- x})

Simplify

= arctan (e^{- x}) - e^{- x}

Add a constant to the solution

= arctan (e^{- x}) - e^{- x} + C

\int \frac{x - x}{x ^{2}} d x

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

\int csc (x) - sin (x) d x

\int \frac{1}{x ^{2} + 12} d x

What is the integral of (e^{-x})/(1+e^{2x)} ?
The integral of (e^{-x})/(1+e^{2x)} is arctan(e^{-x})-e^{-x}+C