What is the integral of (e^u)/((1-e^u)^2) ?

The integral of (e^u)/((1-e^u)^2) is 1/(1-e^u)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^u)/((1-e^u)^2)

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

\frac{1}{1 - e ^{u}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{v ^{2}} d v

= \int - \frac{1}{v ^{2}} d v

Take the constant out:

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

= - \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - (- \frac{1}{v})

Substitute back

v = 1 - e^{u}

= - (- \frac{1}{1 - e ^{u}})

Simplify

= \frac{1}{1 - e ^{u}}

Add a constant to the solution

= \frac{1}{1 - e ^{u}} + C

\frac{d y}{d x} = y^{2} (4 - e^{x})

3 y^{^{''}} + 5 y^{^{'}} - 12 y = 0

\frac{d}{d y} (- 2 e^{x} + 4 x cos (y))

\int 2 x - 8 y d x

\int \frac{x ^{3} - 2 x}{x ^{2} + 3 x + 2} d x

What is the integral of (e^u)/((1-e^u)^2) ?
The integral of (e^u)/((1-e^u)^2) is 1/(1-e^u)+C