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

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

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

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

2 y + e^{y} - e^{- y} - 2 + C

Solution steps

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

Apply Integration By Parts

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

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

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

= - e^{- y} (e^{y} + 1)^{2} - (- 2 (e^{y} + y))

Simplify

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

= 2 y + e^{y} - e^{- y} - 2

Add a constant to the solution

= 2 y + e^{y} - e^{- y} - 2 + C

\int \frac{x ^{3}}{x ^{2} + x - 2} d x

\int \frac{x ^{4}}{x ^{5} - 2} d x

\frac{d}{d x} (\frac{e ^{\frac{1}{x}}}{x ^{2}})

x \to \infty lim (1 - \frac{3}{x})

\frac{d}{d y} (- x y)

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