What is the integral of e^y(e^y+1)^{-2} ?

The integral of e^y(e^y+1)^{-2} is -1/(e^y+1)+C

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integral of e^y(e^y+1)^{-2}

\int e^{y} (e^{y} + 1)^{- 2} d y

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

Solution steps

\int e^{y} (e^{y} + 1)^{- 2} d y

Simplify

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

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = e^{y} + 1

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

Add a constant to the solution

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

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What is the integral of e^y(e^y+1)^{-2} ?
The integral of e^y(e^y+1)^{-2} is -1/(e^y+1)+C