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

Simplify

= ln 1 + e^{- x} + x

Add a constant to the solution

= ln 1 + e^{- x} + x + C

2 x (x^{2} + 3)^{\frac{3}{2}}

y = sin^{2} (x)

\frac{1}{( x ^{4} )}

lo g_{10} (\frac{( 1 + x )}{( 1 - x )})

\frac{1}{y ^{3}}

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