What is the integral of ((u+1))/(u(u+2)) ?

The integral of ((u+1))/(u(u+2)) is 1/2 ln|u|+1/2 ln|u+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((u+1))/(u(u+2))

\int \frac{( u + 1 )}{u ( u + 2 )} d u

\frac{1}{2} ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 2 ∣ + C

Solution steps

\int \frac{u + 1}{u ( u + 2 )} d u

Expand

\frac{u + 1}{u ( u + 2 )} : \frac{u}{u ( u + 2 )} + \frac{1}{u ( u + 2 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{u}{u ( u + 2 )} d u = ln ∣ u + 2 ∣

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

= ln ∣ u + 2 ∣ + \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Simplify

ln ∣ u + 2 ∣ + \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣ : \frac{1}{2} ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 2 ∣ + C

What is the integral of ((u+1))/(u(u+2)) ?
The integral of ((u+1))/(u(u+2)) is 1/2 ln|u|+1/2 ln|u+2|+C