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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = e^{x}

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

Simplify

- (\frac{ln ∣ e ^{x} + 1 ∣}{2} - \frac{ln ∣ e ^{x} - 1 ∣}{2}) : - \frac{1}{2} ln ∣ e^{x} + 1 ∣ + \frac{1}{2} ln ∣ e^{x} - 1 ∣

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

Add a constant to the solution

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

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