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

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

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

\int \frac{e ^{2 y}}{e ^{y} - 1} d y

e^{y} - 1 + ln ∣ e^{y} - 1 ∣ + C

Solution steps

\int \frac{e ^{2 y}}{e ^{y} - 1} d y

Apply u-substitution

: \int \frac{1}{2 ( u - 1 )} d u

= \int \frac{1}{2 ( u - 1 )} 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{1}{u - 1} d u

Apply u-substitution

: \int \frac{2 ( v + 1 )}{v} d v

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

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{v + 1}{v} d v

Expand

\frac{v + 1}{v} : 1 + \frac{1}{v}

= \frac{1}{2} \cdot 2 \cdot \int 1 + \frac{1}{v} d v

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} \cdot 2 (\int 1 d v + \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{2} \cdot 2 (v + ln ∣ v ∣)

Substitute back

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

Simplify

\frac{1}{2} \cdot 2 (e^{2 y} - 1 + ln e^{2 y} - 1) : e^{y} - 1 + ln ∣ e^{y} - 1 ∣

= e^{y} - 1 + ln ∣ e^{y} - 1 ∣

Add a constant to the solution

= e^{y} - 1 + ln ∣ e^{y} - 1 ∣ + C

\int sec^{3} (5 x) d x

y = arcsin (x)

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

2 z

\frac{\partial}{\partial y} (x^{2} \cdot e^{- 2 y})

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