What is the integral of 1/((x^3+3x^2+2x)(x-1)) ?

The integral of 1/((x^3+3x^2+2x)(x-1)) is -1/2 ln|x|+1/2 ln|x+1|-1/6 ln|x+2|+1/6 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 1/((x^3+3x^2+2x)(x-1))

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

- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{3} + 3 x ^{2} + 2 x ) ( x - 1 )} : - \frac{1}{2 x} + \frac{1}{2 ( x + 1 )} - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( x - 1 )}

= \int - \frac{1}{2 x} + \frac{1}{2 ( x + 1 )} - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( 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

= - \int \frac{1}{2 x} d x + \int \frac{1}{2 ( x + 1 )} d x - \int \frac{1}{6 ( x + 2 )} d x + \int \frac{1}{6 ( x - 1 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + C

What is the integral of 1/((x^3+3x^2+2x)(x-1)) ?
The integral of 1/((x^3+3x^2+2x)(x-1)) is -1/2 ln|x|+1/2 ln|x+1|-1/6 ln|x+2|+1/6 ln|x-1|+C