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

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

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integral of 1/2 (e^{2x}-e^{-2x})

\int \frac{1}{2} (e^{2 x} - e^{- 2 x}) d x

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

Solution steps

\int \frac{1}{2} (e^{2 x} - e^{- 2 x}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

e^{2 x} - e^{- 2 x} = (e^{x} + e^{- x}) (e^{x} - e^{- x})

= \frac{1}{2} \cdot \int (e^{x} + e^{- x}) (e^{x} - e^{- x}) d x

Apply u-substitution

: \int u d u

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = e^{x} + e^{- x}

= \frac{1}{2} \cdot \frac{( e ^{x} + e ^{- x} ) ^{2}}{2}

Simplify

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

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

Add a constant to the solution

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

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