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

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

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

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

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

Decimal

0.62011 \dots

Solution steps

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

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

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

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

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

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

Simplify

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

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