What is the integral of (5e^{2x}-e^x)/(e^{2x)-1} ?

The integral of (5e^{2x}-e^x)/(e^{2x)-1} is 5/2 ln|e^{2x}-1|+1/2 ln|e^x+1|-1/2 ln|e^x-1|+C

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integral of (5e^{2x}-e^x)/(e^{2x)-1}

\int \frac{5 e ^{2 x} - e ^{x}}{e ^{2 x} - 1} d x

\frac{5}{2} ln e^{2 x} - 1 + \frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} - 1 ∣ + C

Solution steps

\int \frac{5 e ^{2 x} - e ^{x}}{e ^{2 x} - 1} d x

Expand

\frac{5 e ^{2 x} - e ^{x}}{e ^{2 x} - 1} : \frac{5 e ^{2 x}}{e ^{2 x} - 1} - \frac{e ^{x}}{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{5 e ^{2 x}}{e ^{2 x} - 1} d x - \int \frac{e ^{x}}{e ^{2 x} - 1} d x

\int \frac{5 e ^{2 x}}{e ^{2 x} - 1} d x = \frac{5}{2} ln e^{2 x} - 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 ∣

= \frac{5}{2} ln e^{2 x} - 1 - (- \frac{1}{2} ln ∣ e^{x} + 1 ∣ + \frac{1}{2} ln ∣ e^{x} - 1 ∣)

Simplify

= \frac{5}{2} ln e^{2 x} - 1 + \frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} - 1 ∣

Add a constant to the solution

= \frac{5}{2} ln e^{2 x} - 1 + \frac{1}{2} ln ∣ e^{x} + 1 ∣ - \frac{1}{2} ln ∣ e^{x} - 1 ∣ + C

\frac{\partial}{\partial x} (9 x^{2} y^{2})

\int (x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x}) d x

\int_{2}^{6} 8 d x

x \to 0 lim (3 sin (x) - 1)

\frac{2 x + 5}{x}

What is the integral of (5e^{2x}-e^x)/(e^{2x)-1} ?
The integral of (5e^{2x}-e^x)/(e^{2x)-1} is 5/2 ln|e^{2x}-1|+1/2 ln|e^x+1|-1/2 ln|e^x-1|+C