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

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

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

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

- ln (e + 1) + ln (2) + 1

Decimal Notation

0.37988 \dots

Solution steps

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

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

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

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

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

= - (ln (e + 1) - ln (2)) + 1

Simplify

= - ln (e + 1) + ln (2) + 1

a y^{^{'}} = b x - b y

\frac{d y}{d x} + (5 x) y = 3 x + 5

\int e^{t} + e^{- t} d t

\frac{\partial}{\partial z} (sin (x y))

\frac{\partial}{\partial y} (ln (\frac{y ^{9}}{x ^{7}}))

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