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

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

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

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

\frac{1}{2} ln 1 + e^{- 2 x} + x + C

Solution steps

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

\frac{1}{1 + e ^{- 2 x}} = - \frac{e ^{- 2 x}}{1 + e ^{- 2 x}} + 1

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

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

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

\int 1 d x = x

= - (- \frac{1}{2} ln 1 + e^{- 2 x}) + x

Simplify

= \frac{1}{2} ln 1 + e^{- 2 x} + x

Add a constant to the solution

= \frac{1}{2} ln 1 + e^{- 2 x} + x + C

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