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

3 cos (2 x), (3 - 3 cos (2 x)) x = 0 x = \frac{π}{2}

\int_{0}^{1} π (5 x)^{2} d x

\frac{d}{d x} (5 - 5 x^{2})

6 y^{^{''}} - 6 y^{^{'}} + 75 y = 0

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (2) = 0

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