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

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

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integral of (e^x-1)/(e^x+1)

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

2 ln ∣ e^{x} + 1 ∣ - x + C

Solution steps

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

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

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

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

\int 1 d x = x

= 2 ln ∣ e^{x} + 1 ∣ - x

Add a constant to the solution

= 2 ln ∣ e^{x} + 1 ∣ - x + C

x \to 1 + lim (x^{2} + 1)

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

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

\frac{y}{y - x}

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