What is the integral of (e^u)/((7-e^u)^2) ?

The integral of (e^u)/((7-e^u)^2) is 1/(7-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)/((7-e^u)^2)

\int \frac{e ^{u}}{( 7 - e ^{u} ) ^{2}} d u

\frac{1}{7 - e ^{u}} + C

Solution steps

\int \frac{e ^{u}}{( 7 - 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 = 7 - e^{u}

= - (- \frac{1}{7 - e ^{u}})

Simplify

= \frac{1}{7 - e ^{u}}

Add a constant to the solution

= \frac{1}{7 - e ^{u}} + C

(x^{2} + 1) d y = (x^{3} - 2 x y + x) d x

arccos (x)

\int 5 sin^{4} (r) cos (r) d r

\frac{e ^{x}}{x ^{2}}

\frac{d}{d x} ((x - a)^{3} (x - b))

What is the integral of (e^u)/((7-e^u)^2) ?
The integral of (e^u)/((7-e^u)^2) is 1/(7-e^u)+C