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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x^2-3x+3)/(x^3-2x^2+x)

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

= 3 ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{2}{x - 1}

Add a constant to the solution

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

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