What is the integral from 0 to 1 of (e^x)/(e^x+1) ?

The integral from 0 to 1 of (e^x)/(e^x+1) is ln(e+1)-ln(2)

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integral from 0 to 1 of (e^x)/(e^x+1)

\int_{0}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

ln (e + 1) - ln (2)

Decimal Notation

0.62011 \dots

Solution steps

\int_{0}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

Apply u-substitution

: \int_{2}^{e + 1} \frac{1}{u} d u

= \int_{2}^{e + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{e + 1}

Compute the boundaries:

ln (e + 1) - ln (2)

= ln (e + 1) - ln (2)

\frac{\partial}{\partial x} (a sin (w (x) x) + b cos (w (x) x))

x - x^{\frac{1}{2}}

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

\frac{4}{s ^{2} + 1}

\frac{d y}{d t} = t y

What is the integral from 0 to 1 of (e^x)/(e^x+1) ?
The integral from 0 to 1 of (e^x)/(e^x+1) is ln(e+1)-ln(2)