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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

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

= - \frac{1}{2} \cdot 2 arctan (w)

Substitute back

= - \frac{1}{2} \cdot 2 arctan (e^{\frac{- 2 x}{2}})

Simplify

- \frac{1}{2} \cdot 2 arctan (e^{\frac{- 2 x}{2}}) : - arctan (e^{- x})

= - arctan (e^{- x})

Add a constant to the solution

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

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