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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{1}{8 x + 2 x ^{2}} d x

x \to 2 lim (2 x^{2} + \frac{a}{2} x - 4)

(x^{2} - x + 2)^{3}

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

f (t) = t^{2} e^{, at}

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