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} (x^{2} - 2 x y + y^{2})

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

\int \frac{y + 1}{y - 1} d y

y^{^{''}} - 5 y^{^{'}} = x - 2

3 x^{2} cos (x) cot (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