What is the integral of (xe^{-x})/((1+e^{-x))^2} ?

The integral of (xe^{-x})/((1+e^{-x))^2} is x/(1+e^{-x)}-x-ln|e^{-x}+1|+C

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integral of (xe^{-x})/((1+e^{-x))^2}

\int \frac{x e ^{- x}}{( 1 + e ^{- x} ) ^{2}} d x

\frac{x}{1 + e ^{- x}} - x - ln e^{- x} + 1 + C

Solution steps

\int \frac{x e ^{- x}}{( 1 + e ^{- x} ) ^{2}} d x

Apply u-substitution

: \int \frac{ln ( u )}{( 1 + u ) ^{2}} d u

= \int \frac{ln ( u )}{( 1 + u ) ^{2}} d u

Apply Integration By Parts

: - \frac{ln ( u )}{1 + u} - \int - \frac{1}{u ( 1 + u )} d u

= - \frac{ln ( u )}{1 + u} - \int - \frac{1}{u ( 1 + u )} d u

\int - \frac{1}{u ( 1 + u )} d u = - ln ∣ u ∣ + ln ∣ u + 1 ∣

= - \frac{ln ( u )}{1 + u} - (- ln ∣ u ∣ + ln ∣ u + 1 ∣)

Substitute back

u = e^{- x}

= - \frac{ln ( e ^{- x} )}{1 + e ^{- x}} - (- ln e^{- x} + ln e^{- x} + 1)

Simplify

- \frac{ln ( e ^{- x} )}{1 + e ^{- x}} - (- ln e^{- x} + ln e^{- x} + 1) : \frac{x}{1 + e ^{- x}} - x - ln e^{- x} + 1

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

Add a constant to the solution

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

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What is the integral of (xe^{-x})/((1+e^{-x))^2} ?
The integral of (xe^{-x})/((1+e^{-x))^2} is x/(1+e^{-x)}-x-ln|e^{-x}+1|+C