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

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

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

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

- 2 ln ∣ 1 - e^{x} ∣ + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - 2 ln ∣ 1 - e^{x} ∣ + x

Add a constant to the solution

= - 2 ln ∣ 1 - e^{x} ∣ + x + C

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