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

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

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

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

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

Solution steps

\int \frac{1}{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

\int_{0}^{\infty} e^{- 2 x} d x

n = 0 \sum \infty \frac{n ^{3}}{e ^{n}}

\int - 2 x^{3} y^{- 3} - x^{2} y^{- 2} d y

\int_{0}^{1} 2 π (x + 8) (x^{2}) d x

\frac{y , ( x + y )}{( x ^{2} + y ^{2} + x y + 3 )}

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