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

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

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

\int (\frac{e ^{x}}{e ^{x} + 1}) d x

ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = e^{x} + 1

= ln ∣ e^{x} + 1 ∣

Add a constant to the solution

= ln ∣ e^{x} + 1 ∣ + C

\frac{\partial}{\partial x} (y cos (x lo g_{e} (y)))

x \to \infty lim (\frac{2}{0})

\frac{d y}{d t} = - 3 (y - 5)

x \to \infty lim (\frac{( \frac{x + 1}{x ^{3} + 2} )}{( \frac{x}{x ^{3}} )})

x \to 5 + lim (\frac{8}{x - 5})

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