What is the integral from 0 to 4 of (e^x)/(4+e^x) ?

The integral from 0 to 4 of (e^x)/(4+e^x) is ln(4+e^4)-ln(5)

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integral from 0 to 4 of (e^x)/(4+e^x)

\int_{0}^{4} \frac{e ^{x}}{4 + e ^{x}} d x

ln (4 + e^{4}) - ln (5)

Decimal

2.46126 \dots

Solution steps

\int_{0}^{4} \frac{e ^{x}}{4 + e ^{x}} d x

Apply u-substitution

: \int_{5}^{4 + e^{4}} \frac{1}{u} d u

= \int_{5}^{4 + e^{4}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{5}^{4 + e^{4}}

Compute the boundaries:

ln (4 + e^{4}) - ln (5)

= ln (4 + e^{4}) - ln (5)

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What is the integral from 0 to 4 of (e^x)/(4+e^x) ?
The integral from 0 to 4 of (e^x)/(4+e^x) is ln(4+e^4)-ln(5)