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

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

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

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

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

Solution steps

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

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 t} - 1

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

Simplify

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

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

Add a constant to the solution

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

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