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

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

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

\int_{1}^{- 1} \frac{e ^{x}}{e ^{x} + 1} d x

- 1

Solution steps

\int_{1}^{- 1} \frac{e ^{x}}{e ^{x} + 1} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} \frac{e ^{x}}{e ^{x} + 1} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{\frac{1}{e} + 1}^{e + 1}

Compute the boundaries:

ln (e + 1) - ln (\frac{1}{e} + 1)

= - (ln (e + 1) - ln (\frac{1}{e} + 1))

Simplify

= - 1

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