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)

integral

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

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{e ^{- x}}{( 1 + e ^{- x} )}

x sin^{2} (x)

\frac{x ^{2}}{9 - x ^{2}}

\frac{x}{( x + 1 ) ^{3}}

y = x^{3} (e^{3})^{4}

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