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

The integral of (2-e^{-x})^2 is 4x+4e^{-x}-1/2 e^{-2x}+C

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

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

4 x + 4 e^{- x} - \frac{1}{2} e^{- 2 x} + C

Solution steps

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

Expand

(2 - e^{- x})^{2} : 4 - 4 e^{- x} + e^{- 2 x}

= \int 4 - 4 e^{- x} + e^{- 2 x} d x

Apply the Sum Rule:

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

= \int 4 d x - \int 4 e^{- x} d x + \int e^{- 2 x} d x

\int 4 d x = 4 x

\int 4 e^{- x} d x = - 4 e^{- x}

\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x}

= 4 x - (- 4 e^{- x}) - \frac{1}{2} e^{- 2 x}

Simplify

= 4 x + 4 e^{- x} - \frac{1}{2} e^{- 2 x}

Add a constant to the solution

= 4 x + 4 e^{- x} - \frac{1}{2} e^{- 2 x} + C

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What is the integral of (2-e^{-x})^2 ?
The integral of (2-e^{-x})^2 is 4x+4e^{-x}-1/2 e^{-2x}+C