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

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

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

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

ln (1 + e) - ln (2)

Decimal Notation

0.62011 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{1 + e}

Compute the boundaries:

ln (1 + e) - ln (2)

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

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\int 2 x^{4} ln (x) d x

\frac{d}{d x} ((4 x - 5) ln (2 x - 3))

\int \frac{2 x + 1}{x ^{2} + 2 x + 2} d x

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