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

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

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

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

- \frac{1}{2} e^{- 2 x} - 2 e^{- x} + x + C

Solution steps

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

Apply u-substitution

= \int - e^{2 u} - 2 e^{u} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int e^{2 u} d u - \int 2 e^{u} d u - \int 1 d u

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

\int 2 e^{u} d u = 2 e^{u}

\int 1 d u = u

= - \frac{1}{2} e^{2 u} - 2 e^{u} - u

Substitute back

u = - x

= - \frac{1}{2} e^{2 (- x)} - 2 e^{- x} - (- x)

Simplify

= - \frac{1}{2} e^{- 2 x} - 2 e^{- x} + x

Add a constant to the solution

= - \frac{1}{2} e^{- 2 x} - 2 e^{- x} + x + C

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