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

The integral of 1/((e^x+1)) is -ln|e^x+1|+x+C

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integral of 1/((e^x+1))

\int \frac{1}{( e ^{x} + 1 )} d x

- ln ∣ e^{x} + 1 ∣ + x + C

Solution steps

\int \frac{1}{e ^{x} + 1} d x

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

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

\int \frac{e ^{x}}{e ^{x} + 1} d x = ln ∣ e^{x} + 1 ∣

\int 1 d x = x

= - ln ∣ e^{x} + 1 ∣ + x

Add a constant to the solution

= - ln ∣ e^{x} + 1 ∣ + x + C

\frac{d}{d x} (\frac{1 + sin ( x )}{cos ( x )})

y^{^{'}} - \frac{4 y}{x} = - \frac{2}{x ^{2}}

\frac{- 6 x}{x ^{2} + 1}

\frac{d}{d y} (2 y^{3})

\int \frac{x ^{3} - 2}{x} d x

What is the integral of 1/((e^x+1)) ?
The integral of 1/((e^x+1)) is -ln|e^x+1|+x+C