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

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

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

integral

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

- ln ∣ 1 + e^{x} ∣ + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - ln ∣ 1 + e^{x} ∣ + x

Add a constant to the solution

= - ln ∣ 1 + e^{x} ∣ + x + C

\frac{( e ^{- x} ) ^{2}}{x}

\frac{1}{( y ^{2} + 1 )}

\frac{x}{( e ^{x} ) ^{2}}

(e^{x} + 2)^{2}

\frac{2 - x}{1 - x}

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