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

The integral from 0 to ln(2) of 1/(1+e^x) is -ln(3)+2ln(2)

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

\int_{0}^{l n (2)} \frac{1}{1 + e ^{x}} d x

- ln (3) + 2 ln (2)

Decimal Notation

0.28768 \dots

Solution steps

\int_{0}^{l n (2)} \frac{1}{1 + e ^{x}} d x

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

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

Apply the Sum Rule:

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

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

\int_{0}^{l n (2)} \frac{e ^{x}}{1 + e ^{x}} d x = ln (3) - ln (2)

\int_{0}^{l n (2)} 1 d x = ln (2)

= - (ln (3) - ln (2)) + ln (2)

Simplify

= - ln (3) + 2 ln (2)

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What is the integral from 0 to ln(2) of 1/(1+e^x) ?
The integral from 0 to ln(2) of 1/(1+e^x) is -ln(3)+2ln(2)