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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{ln ( 2 ) u} d u

= \int \frac{1}{ln ( 2 ) u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 2 )} ln ∣ u ∣

Substitute back

u = 2^{x} + 1

= \frac{1}{ln ( 2 )} ln ∣ 2^{x} + 1 ∣

Add a constant to the solution

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

y = x - 1

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

\frac{d}{d x} (- cos (2 x) \cdot 2)

\frac{\partial}{\partial y} (y^{2} e^{x y - 3})

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