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

解答

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

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

求解步骤

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

使用换元积分法

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

使用分布积分法

= - \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 ∣)

u = e^{x}

代回

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

化简

- \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 ∣

解答补常数

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

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