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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

(\frac{x ^{2} - 2}{5 x ^{3} + 6})^{^{'}}

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 5, y^{^{'}} (0) = 10

\frac{\partial}{\partial x} (tanh (x))

x \to 0 lim (\frac{tan ( 5 x )}{x})

\int \frac{1}{x ^{3}} - x d x

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