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

(\frac{1}{2})^{x} - x

\int_{- 1}^{1} \frac{3}{4} x^{2} (1 - x^{2}) d x

\int - 5 tan^{3} (x) d x

f (3)

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

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