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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