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

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

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

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

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Apply u-substitution

= - \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 = - 1 + e^{2 x}

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

Add a constant to the solution

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

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