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

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

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

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

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

Decimal

1

Solution steps

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

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

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

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

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

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

Simplify

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

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