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

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

integral from 0 to 1 of 1/(1+e^{x-1)}

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

- ln (2) + ln (1 + \frac{1}{e}) + 1

Decimal

0.62011 \dots

Solution steps

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

Apply u-substitution

= \int_{- 1}^{0} \frac{1}{1 + e ^{u}} d u

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

= \int_{- 1}^{0} - \frac{e ^{u}}{1 + e ^{u}} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{0} \frac{e ^{u}}{1 + e ^{u}} d u + \int_{- 1}^{0} 1 d u

\int_{- 1}^{0} \frac{e ^{u}}{1 + e ^{u}} d u = ln (2) - ln (1 + \frac{1}{e})

\int_{- 1}^{0} 1 d u = 1

= - (ln (2) - ln (1 + \frac{1}{e})) + 1

Simplify

= - ln (2) + ln (1 + \frac{1}{e}) + 1

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