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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ^{2} ( u - 1 )} : - \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u - 1}

= \int - \frac{1}{u} - \frac{1}{u ^{2}} + \frac{1}{u - 1} d u

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{1}{u} d u - \int \frac{1}{u ^{2}} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

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

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

Substitute back

u = e^{x} + 1

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

Simplify

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

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

Add a constant to the solution

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

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