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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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