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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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