What is the integral from 0 to 2 of (e^x)/(10+e^x) ?

The integral from 0 to 2 of (e^x)/(10+e^x) is ln(10+e^2)-ln(11)

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integral from 0 to 2 of (e^x)/(10+e^x)

\int_{0}^{2} \frac{e ^{x}}{10 + e ^{x}} d x

ln (10 + e^{2}) - ln (11)

Decimal

0.45794 \dots

Solution steps

\int_{0}^{2} \frac{e ^{x}}{10 + e ^{x}} d x

Apply u-substitution

: \int_{11}^{10 + e^{2}} \frac{1}{u} d u

= \int_{11}^{10 + e^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{11}^{10 + e^{2}}

Compute the boundaries:

ln (10 + e^{2}) - ln (11)

= ln (10 + e^{2}) - ln (11)

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What is the integral from 0 to 2 of (e^x)/(10+e^x) ?
The integral from 0 to 2 of (e^x)/(10+e^x) is ln(10+e^2)-ln(11)