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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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