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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} - 1 ) ^{2}}{2 u ^{3}} d u

= \int \frac{( u ^{2} - 1 ) ^{2}}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{( u ^{2} - 1 ) ^{2}}{u ^{3}} d u

Expand

\frac{( u ^{2} - 1 ) ^{2}}{u ^{3}} : u - \frac{2}{u} + \frac{1}{u ^{3}}

= \frac{1}{2} \cdot \int u - \frac{2}{u} + \frac{1}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u d u - \int \frac{2}{u} d u + \int \frac{1}{u ^{3}} d u)

\int u d u = \frac{u ^{2}}{2}

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= \frac{1}{2} (\frac{u ^{2}}{2} - 2 ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Substitute back

u = e^{2 x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} = k y

\frac{\partial}{\partial r} (r^{2})

x \to - \infty lim (x^{3})

\frac{\partial}{\partial x} (\frac{( x \cdot y )}{( x + y )})

\int y t^{3} - 1.5 y d t

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