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

The integral of (e^{x/a}-e^{(-x)/a})^2 is (ae^{(2x)/a})/(2)-2x-a/(2e^{\frac{2x){a}}}+C

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

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

\frac{a e ^{\frac{2 x}{a}}}{2} - 2 x - \frac{a}{2 e ^{\frac{2 x}{a}}} + C

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Take the constant out:

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

= a \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}}

= a \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

= a (\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}}

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

Substitute back

u = e^{\frac{x}{a}}

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

Simplify

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

= \frac{a e ^{\frac{2 x}{a}}}{2} - 2 x - \frac{a}{2 e ^{\frac{2 x}{a}}}

Add a constant to the solution

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

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What is the integral of (e^{x/a}-e^{(-x)/a})^2 ?
The integral of (e^{x/a}-e^{(-x)/a})^2 is (ae^{(2x)/a})/(2)-2x-a/(2e^{\frac{2x){a}}}+C