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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= - \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

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