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

\int \frac{1}{( x ln ( x ) )} d x

\int x^{2 n - 1} d x

sin (3 x)

\int \frac{( 2 x + 3 )}{x ^{2} + 2 x + 5} d x

\int \frac{1}{5 x lo g _{3} ( x )} d x

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