$\int (e^{x} + e^{- x}) (e^{2 x} - e^{- 2 x}) d x = . . . . + c$ .

\frac{{(e^{x} + e^{- x})}^{3}}{3}

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\frac{{(e^{3 x} + e^{- 3 x})}}{3}

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\frac{{(e^{x} - e^{- x})}^{3}}{3}

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\frac{{(e^{3 x} - e^{- 3 x})}}{3}

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Solution

The correct option is A $\frac{{(e^{x} + e^{- x})}^{3}}{3}$
$\int (e^{x} + e^{- x}) (e^{2 x} - e^{- 2 x}) = \int (e^{3 x} + e^{x} - e^{- x} - e^{- 3 x})$
$\int (e^{3 x} + e^{x} - e^{- x} - e^{- 3 x}) = \frac{e^{3 x}}{3} + \frac{e^{x}}{1} - \frac{e^{- x}}{- 1} - \frac{e^{- 3 x}}{- 3}$
$= \frac{e^{3 x}}{3} + \frac{e^{x}}{1} + \frac{e^{- x}}{1} + \frac{e^{- 3 x}}{3} = \frac{1}{3} (e^{3 x} + 3 e^{x} + 3 e^{- x} + e^{- 3 x}) = \frac{1}{3} (e^{x} + e^{- x})^{3}$

as $(a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} = (a + b)^{3}$

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