What is the integral from 0 to 5 of (e^x)/(2+e^x) ?

The integral from 0 to 5 of (e^x)/(2+e^x) is ln(2+e^5)-ln(3)

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integral from 0 to 5 of (e^x)/(2+e^x)

\int_{0}^{5} \frac{e ^{x}}{2 + e ^{x}} d x

ln (2 + e^{5}) - ln (3)

Decimal

3.91477 \dots

Solution steps

\int_{0}^{5} \frac{e ^{x}}{2 + e ^{x}} d x

Apply u-substitution

: \int_{3}^{2 + e^{5}} \frac{1}{u} d u

= \int_{3}^{2 + e^{5}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{2 + e^{5}}

Compute the boundaries:

ln (2 + e^{5}) - ln (3)

= ln (2 + e^{5}) - ln (3)

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What is the integral from 0 to 5 of (e^x)/(2+e^x) ?
The integral from 0 to 5 of (e^x)/(2+e^x) is ln(2+e^5)-ln(3)