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

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

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integral of (2^x-1)/(2^x+1)

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

\frac{1}{ln ( 2 )} ln ∣ 2^{x} + 1 ∣ - (- \frac{1}{ln ( 2 )} ln ∣ 2^{x} + 1 ∣ + x) : \frac{2}{ln ( 2 )} ln ∣ 2^{x} + 1 ∣ - x

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

Add a constant to the solution

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

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