What is the integral of 4/((x+2)(x-1)) ?

The integral of 4/((x+2)(x-1)) is 4(-1/3 ln|x+2|+1/3 ln|x-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 4/((x+2)(x-1))

\int \frac{4}{( x + 2 ) ( x - 1 )} d x

4 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣) + C

Solution steps

\int \frac{4}{( x + 2 ) ( x - 1 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x + 2 ) ( x - 1 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x - 1 )} : - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( x - 1 )}

= 4 \cdot \int - \frac{1}{3 ( x + 2 )} + \frac{1}{3 ( x - 1 )} d x

Apply the Sum Rule:

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

= 4 (- \int \frac{1}{3 ( x + 2 )} d x + \int \frac{1}{3 ( x - 1 )} d x)

\int \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} ln ∣ x + 2 ∣

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

= 4 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣)

Add a constant to the solution

= 4 (- \frac{1}{3} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x - 1 ∣) + C

What is the integral of 4/((x+2)(x-1)) ?
The integral of 4/((x+2)(x-1)) is 4(-1/3 ln|x+2|+1/3 ln|x-1|)+C