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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

x \to 0 - lim (1 + \frac{1}{x})

\int_{- \infty}^{6} 10 x e^{2 x} d x

\frac{d}{d x} (ln (x x^{2} - 2))

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 sin (2 t)

\int tan^{2} (3 x) sec^{4} (3 x) d x

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