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

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

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integral from 1 to 2 of (e^x)/(e^x+e)

\int_{1}^{2} \frac{e ^{x}}{e ^{x} + e} d x

ln (e^{2} + e) - ln (2 e)

Decimal

0.62011 \dots

Solution steps

\int_{1}^{2} \frac{e ^{x}}{e ^{x} + e} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2 e}^{e^{2} + e}

Compute the boundaries:

ln (e^{2} + e) - ln (2 e)

= ln (e^{2} + e) - ln (2 e)

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