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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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

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