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

The integral of (e^{4x}+e^{-4x})^2 is 1/4 (1/2 e^{8x}+8x-1/(2e^{8x)})+C

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integral of (e^{4x}+e^{-4x})^2

\int (e^{4 x} + e^{- 4 x})^{2} d x

\frac{1}{4} (\frac{1}{2} e^{8 x} + 8 x - \frac{1}{2 e ^{8 x}}) + C

Solution steps

\int (e^{4 x} + e^{- 4 x})^{2} d x

Apply u-substitution

: \int \frac{( u ^{2} + 1 ) ^{2}}{4 u ^{3}} d u

= \int \frac{( u ^{2} + 1 ) ^{2}}{4 u ^{3}} d u

Take the constant out:

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

= \frac{1}{4} \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}{4} \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}{4} (\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}{4} (\frac{u ^{2}}{2} + 2 ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Substitute back

u = e^{4 x}

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

Simplify

\frac{1}{4} (\frac{( e ^{4 x} ) ^{2}}{2} + 2 ln e^{4 x} - \frac{1}{2 ( e ^{4 x} ) ^{2}}) : \frac{1}{4} (\frac{1}{2} e^{8 x} + 8 x - \frac{1}{2 e ^{8 x}})

= \frac{1}{4} (\frac{1}{2} e^{8 x} + 8 x - \frac{1}{2 e ^{8 x}})

Add a constant to the solution

= \frac{1}{4} (\frac{1}{2} e^{8 x} + 8 x - \frac{1}{2 e ^{8 x}}) + C

f (x) = - \frac{10}{x ^{2}}

\int 8 - 2 x^{2} d x

\frac{\partial}{\partial y} (4 arctan (\frac{x}{y}))

\frac{d}{d x} (\frac{x ^{2} - 2 x - 48}{x + 6})

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

What is the integral of (e^{4x}+e^{-4x})^2 ?
The integral of (e^{4x}+e^{-4x})^2 is 1/4 (1/2 e^{8x}+8x-1/(2e^{8x)})+C