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

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

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

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

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

Solution steps

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

Expand

\frac{e ^{- x} - 1}{e ^{- x} + 1} : \frac{e ^{- x}}{e ^{- x} + 1} - \frac{1}{e ^{- x} + 1}

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

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

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

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

Simplify

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

Add a constant to the solution

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

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