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

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

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integral of 1/(1+e^y)

\int \frac{1}{1 + e ^{y}} d y

- ln ∣ 1 + e^{y} ∣ + y + C

Solution steps

\int \frac{1}{1 + e ^{y}} d y

\frac{1}{1 + e ^{y}} = - \frac{e ^{y}}{1 + e ^{y}} + 1

= \int - \frac{e ^{y}}{1 + e ^{y}} + 1 d y

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{e ^{y}}{1 + e ^{y}} d y + \int 1 d y

\int \frac{e ^{y}}{1 + e ^{y}} d y = ln ∣ 1 + e^{y} ∣

\int 1 d y = y

= - ln ∣ 1 + e^{y} ∣ + y

Add a constant to the solution

= - ln ∣ 1 + e^{y} ∣ + y + C

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