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

= \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

\frac{d y}{d x} = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

\frac{10}{s ( s ^{2} + 8 s + 25 )}

\int x e^{- \frac{x}{5}} d x

(\frac{1}{x ^{2} + 1})^{^{'}}

\frac{d y}{d x}, y^{2} - x = 0

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