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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (\frac{1}{3} (e^{y} - 1)^{\frac{3}{2}} + e^{y} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{u - 1} d u

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

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \int v^{2} + 1 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

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

\int v^{2} d v = \frac{v ^{3}}{3}

\int 1 d v = v

= 2 (\frac{v ^{3}}{3} + v)

Substitute back

= 2 (\frac{( e ^{y} - 1 ) ^{3}}{3} + e^{y} - 1)

Simplify

2 (\frac{( e ^{y} - 1 ) ^{3}}{3} + e^{y} - 1) : 2 (\frac{1}{3} (e^{y} - 1)^{\frac{3}{2}} + e^{y} - 1)

= 2 (\frac{1}{3} (e^{y} - 1)^{\frac{3}{2}} + e^{y} - 1)

Add a constant to the solution

= 2 (\frac{1}{3} (e^{y} - 1)^{\frac{3}{2}} + e^{y} - 1) + C

\frac{d y}{d x} - x y^{2} = 2 x y

\int t cos^{5} (t^{2}) d t

y = 1 + 2 e^{8 x}

n = 1 \sum \infty \frac{1}{7 n ^{2} + n + 6}

\frac{d}{d x} (- c e^{- x})

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