What is the integral of (e^{4x}-e^{-4x})^2 ?

The integral of (e^{4x}-e^{-4x})^2 is 1/8 e^{8x}-2x-1/8 e^{-8x}+C

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integral of (e^{4x}-e^{-4x})^2

\int (e^{4 x} - e^{- 4 x})^{2} d x

\frac{1}{8} e^{8 x} - 2 x - \frac{1}{8} e^{- 8 x} + C

Solution steps

\int (e^{4 x} - e^{- 4 x})^{2} d x

Expand

(e^{4 x} - e^{- 4 x})^{2} : e^{8 x} - 2 + e^{- 8 x}

= \int e^{8 x} - 2 + e^{- 8 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^{8 x} d x - \int 2 d x + \int e^{- 8 x} d x

\int e^{8 x} d x = \frac{1}{8} e^{8 x}

\int 2 d x = 2 x

\int e^{- 8 x} d x = - \frac{1}{8} e^{- 8 x}

= \frac{1}{8} e^{8 x} - 2 x - \frac{1}{8} e^{- 8 x}

Add a constant to the solution

= \frac{1}{8} e^{8 x} - 2 x - \frac{1}{8} e^{- 8 x} + C

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What is the integral of (e^{4x}-e^{-4x})^2 ?
The integral of (e^{4x}-e^{-4x})^2 is 1/8 e^{8x}-2x-1/8 e^{-8x}+C