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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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