What is the integral of (e^{3x}+e^{-x})^2e^{x+2} ?

The integral of (e^{3x}+e^{-x})^2e^{x+2} is 1/7 e^{7x+2}+2/3 e^{3x+2}-e^{-x+2}+C

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integral of (e^{3x}+e^{-x})^2e^{x+2}

\int (e^{3 x} + e^{- x})^{2} e^{x + 2} d x

\frac{1}{7} e^{7 x + 2} + \frac{2}{3} e^{3 x + 2} - e^{- x + 2} + C

Solution steps

\int (e^{3 x} + e^{- x})^{2} e^{x + 2} d x

Expand

(e^{3 x} + e^{- x})^{2} e^{x + 2} : e^{7 x + 2} + 2 e^{3 x + 2} + e^{- x + 2}

= \int e^{7 x + 2} + 2 e^{3 x + 2} + e^{- x + 2} 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^{7 x + 2} d x + \int 2 e^{3 x + 2} d x + \int e^{- x + 2} d x

\int e^{7 x + 2} d x = \frac{1}{7} e^{7 x + 2}

\int 2 e^{3 x + 2} d x = \frac{2}{3} e^{3 x + 2}

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

= \frac{1}{7} e^{7 x + 2} + \frac{2}{3} e^{3 x + 2} - e^{- x + 2}

Add a constant to the solution

= \frac{1}{7} e^{7 x + 2} + \frac{2}{3} e^{3 x + 2} - e^{- x + 2} + C

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What is the integral of (e^{3x}+e^{-x})^2e^{x+2} ?
The integral of (e^{3x}+e^{-x})^2e^{x+2} is 1/7 e^{7x+2}+2/3 e^{3x+2}-e^{-x+2}+C