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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 4 d x = 4 x

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

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

Add a constant to the solution

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

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