What is the integral of (e^{3x}+e^{-3x})^2 ?

The integral of (e^{3x}+e^{-3x})^2 is 1/3 (1/2 e^{6x}+6x-1/(2e^{6x)})+C

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integral of (e^{3x}+e^{-3x})^2

\int (e^{3 x} + e^{- 3 x})^{2} d x

\frac{1}{3} (\frac{1}{2} e^{6 x} + 6 x - \frac{1}{2 e ^{6 x}}) + C

Solution steps

\int (e^{3 x} + e^{- 3 x})^{2} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{3} \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}{3} \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}{3} (\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}{3} (\frac{u ^{2}}{2} + 2 ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Substitute back

u = e^{3 x}

= \frac{1}{3} (\frac{( e ^{3 x} ) ^{2}}{2} + 2 ln e^{3 x} - \frac{1}{2 ( e ^{3 x} ) ^{2}})

Simplify

\frac{1}{3} (\frac{( e ^{3 x} ) ^{2}}{2} + 2 ln e^{3 x} - \frac{1}{2 ( e ^{3 x} ) ^{2}}) : \frac{1}{3} (\frac{1}{2} e^{6 x} + 6 x - \frac{1}{2 e ^{6 x}})

= \frac{1}{3} (\frac{1}{2} e^{6 x} + 6 x - \frac{1}{2 e ^{6 x}})

Add a constant to the solution

= \frac{1}{3} (\frac{1}{2} e^{6 x} + 6 x - \frac{1}{2 e ^{6 x}}) + C

\frac{d}{d x} (ln (4 - 3 x^{3}))

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

t \cdot y^{^{'}} + 2 \cdot y = t^{2} - t + 1

f (x) = e^{- 3 x} sin (4 x^{2})

\int e^{9 - 4 x} d x

What is the integral of (e^{3x}+e^{-3x})^2 ?
The integral of (e^{3x}+e^{-3x})^2 is 1/3 (1/2 e^{6x}+6x-1/(2e^{6x)})+C