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

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

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integral from 0 to ln(3) of 1/(1+e^x)

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

- ln (2) + ln (3)

Decimal

0.40546 \dots

Solution steps

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

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

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

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

\int_{0}^{l n (3)} 1 d x = ln (3)

= - ln (2) + ln (3)

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