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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{u + 1}{2 u} 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} d u

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (u + ln ∣ u ∣)

Substitute back

u = e^{2 x} - 1

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

Add a constant to the solution

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

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