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

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

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

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

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

Solution steps

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

Expand

e^{2 x} (e^{x} - e^{- x})^{2} : e^{4 x} - 2 e^{2 x} + 1

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

\int e^{4 x} d x = \frac{1}{4} e^{4 x}

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

\int 1 d x = x

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

Add a constant to the solution

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

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