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

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

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

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

e^{x} + 1 - x + C

Solution steps

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

Expand

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

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

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

= e^{x} + 1 - ln ∣ e^{x} + 1 ∣ - (- ln ∣ e^{x} + 1 ∣ + x)

Simplify

= e^{x} + 1 - x

Add a constant to the solution

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

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