What is the integral from 0 to 2 of (e^x)/(e^x+1) ?

The integral from 0 to 2 of (e^x)/(e^x+1) is ln(e^2+1)-ln(2)

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integral from 0 to 2 of (e^x)/(e^x+1)

\int_{0}^{2} \frac{e ^{x}}{e ^{x} + 1} d x

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

Decimal Notation

1.43378 \dots

Solution steps

\int_{0}^{2} \frac{e ^{x}}{e ^{x} + 1} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{e^{2} + 1}

Compute the boundaries:

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

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

\int_{1}^{x} (2 x y + 3 y^{2}) d y

\int_{0}^{t} cosh (2 u) d u

\int_{- 2}^{2} \frac{32}{x ^{2} + 4} - x^{2} d x

\int_{0}^{2} \frac{1}{( 1 - x ^{2} ) x ^{2}} d x

\int_{- 2}^{1} (x^{2} + 1) d x

What is the integral from 0 to 2 of (e^x)/(e^x+1) ?
The integral from 0 to 2 of (e^x)/(e^x+1) is ln(e^2+1)-ln(2)