What is the integral from 0 to 4 of 1/(e^x+1) ?

The integral from 0 to 4 of 1/(e^x+1) is -ln(e^4+1)+ln(2)+4

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integral from 0 to 4 of 1/(e^x+1)

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

- ln (e^{4} + 1) + ln (2) + 4

Decimal Notation

0.67499 \dots

Solution steps

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

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

= \int_{0}^{4} - \frac{e ^{x}}{e ^{x} + 1} + 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}^{4} \frac{e ^{x}}{e ^{x} + 1} d x + \int_{0}^{4} 1 d x

\int_{0}^{4} \frac{e ^{x}}{e ^{x} + 1} d x = ln (e^{4} + 1) - ln (2)

\int_{0}^{4} 1 d x = 4

= - (ln (e^{4} + 1) - ln (2)) + 4

Simplify

= - ln (e^{4} + 1) + ln (2) + 4

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What is the integral from 0 to 4 of 1/(e^x+1) ?
The integral from 0 to 4 of 1/(e^x+1) is -ln(e^4+1)+ln(2)+4