What is the integral of (e^{4x})/((e^{2x)-1)^3} ?

The integral of (e^{4x})/((e^{2x)-1)^3} is -(2e^{2x}-1)/(4(e^x+1)(e^x-1)(e^{2x)-1)}+C

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integral of (e^{4x})/((e^{2x)-1)^3}

\int \frac{e ^{4 x}}{( e ^{2 x} - 1 ) ^{3}} d x

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

Solution steps

\int \frac{e ^{4 x}}{( e ^{2 x} - 1 ) ^{3}} d x

Apply u-substitution

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

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

Expand

\frac{u + 1}{u ^{3}} : \frac{1}{u ^{2}} + \frac{1}{u ^{3}}

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

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

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

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

Substitute back

u = e^{2 x} - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (\frac{3}{x}))

x \to 2 + lim (\frac{1}{( x - 2 ) ^{3}})

x \to 0 lim (\frac{x}{x ^{2} - 4 x})

\int \frac{1}{( a x + b ) ^{2}} d x

x \to 3 lim (\frac{1}{∣ x - 3 ∣})

What is the integral of (e^{4x})/((e^{2x)-1)^3} ?
The integral of (e^{4x})/((e^{2x)-1)^3} is -(2e^{2x}-1)/(4(e^x+1)(e^x-1)(e^{2x)-1)}+C