What is the integral of (5x^2-3x+2)/(x^3-2x^2) ?

The integral of (5x^2-3x+2)/(x^3-2x^2) is ln|x|+4ln|x-2|+1/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (5x^2-3x+2)/(x^3-2x^2)

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x

Expand

\frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} : \frac{5 x ^{2}}{x ^{3} - 2 x ^{2}} - \frac{3 x}{x ^{3} - 2 x ^{2}} + \frac{2}{x ^{3} - 2 x ^{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{5 x ^{2}}{x ^{3} - 2 x ^{2}} d x - \int \frac{3 x}{x ^{3} - 2 x ^{2}} d x + \int \frac{2}{x ^{3} - 2 x ^{2}} d x

\int \frac{5 x ^{2}}{x ^{3} - 2 x ^{2}} d x = 5 ln ∣ x - 2 ∣

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

\int \frac{2}{x ^{3} - 2 x ^{2}} d x = - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣

= 5 ln ∣ x - 2 ∣ - 3 (- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣

Simplify

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

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

Add a constant to the solution

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

What is the integral of (5x^2-3x+2)/(x^3-2x^2) ?
The integral of (5x^2-3x+2)/(x^3-2x^2) is ln|x|+4ln|x-2|+1/x+C