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

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

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integral from 0 to 2 of (e^x)/(3+e^x)

\int_{0}^{2} \frac{e ^{x}}{3 + e ^{x}} d x

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

Decimal

0.95445 \dots

Solution steps

\int_{0}^{2} \frac{e ^{x}}{3 + e ^{x}} d x

Apply u-substitution

: \int_{4}^{3 + e^{2}} \frac{1}{u} d u

= \int_{4}^{3 + e^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{4}^{3 + e^{2}}

Compute the boundaries:

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

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

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