What is the integral from 2 to 8 of (e^x)/(1+e^x) ?

The integral from 2 to 8 of (e^x)/(1+e^x) is ln(1+e^8)-ln(1+e^2)

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integral from 2 to 8 of (e^x)/(1+e^x)

\int_{2}^{8} \frac{e ^{x}}{1 + e ^{x}} d x

ln (1 + e^{8}) - ln (1 + e^{2})

Decimal

5.87340 \dots

Solution steps

\int_{2}^{8} \frac{e ^{x}}{1 + e ^{x}} d x

Apply u-substitution

: \int_{1 + e^{2}}^{1 + e^{8}} \frac{1}{u} d u

= \int_{1 + e^{2}}^{1 + e^{8}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1 + e^{2}}^{1 + e^{8}}

Compute the boundaries:

ln (1 + e^{8}) - ln (1 + e^{2})

= ln (1 + e^{8}) - ln (1 + e^{2})

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\int_{0}^{3} - x^{2} + 4 x d x

What is the integral from 2 to 8 of (e^x)/(1+e^x) ?
The integral from 2 to 8 of (e^x)/(1+e^x) is ln(1+e^8)-ln(1+e^2)