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

2 x (4 - x^{2})

\frac{1}{( x ^{- 0} 0.5 + x ^{0.5} )}

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

\frac{ln ( 1 + x ^{2} )}{x ^{2}}