What is the integral of (x^2)/(x^2-5x+6) ?

The integral of (x^2)/(x^2-5x+6) is 4(5/8 ln|4x^2-20x+24|-13/8 ln|2x-4|+13/8 ln|2x-6|)+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2)/(x^2-5x+6)

\int \frac{x ^{2}}{x ^{2} - 5 x + 6} d x

4 (\frac{5}{8} ln 4 x^{2} - 20 x + 24 - \frac{13}{8} ln ∣ 2 x - 4 ∣ + \frac{13}{8} ln ∣ 2 x - 6 ∣) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 5 x + 6} d x

\frac{x ^{2}}{x ^{2} - 5 x + 6} = \frac{5 x - 6}{x ^{2} - 5 x + 6} + 1

= \int \frac{5 x - 6}{x ^{2} - 5 x + 6} + 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{5 x - 6}{x ^{2} - 5 x + 6} d x + \int 1 d x

\int \frac{5 x - 6}{x ^{2} - 5 x + 6} d x = 4 (5 (\frac{1}{8} ln 4 x^{2} - 20 x + 24 - \frac{5}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + 3 (\frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣))

\int 1 d x = x

= 4 (5 (\frac{1}{8} ln 4 x^{2} - 20 x + 24 - \frac{5}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + 3 (\frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣)) + x

Expand

5 (\frac{1}{8} ln 4 x^{2} - 20 x + 24 - \frac{5}{8} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + 3 (\frac{1}{2} ln ∣ 2 x - 4 ∣ - \frac{1}{2} ln ∣ 2 x - 6 ∣) : \frac{5}{8} ln 4 x^{2} - 20 x + 24 - \frac{13}{8} ln ∣ 2 x - 4 ∣ + \frac{13}{8} ln ∣ 2 x - 6 ∣

= 4 (\frac{5}{8} ln 4 x^{2} - 20 x + 24 - \frac{13}{8} ln ∣ 2 x - 4 ∣ + \frac{13}{8} ln ∣ 2 x - 6 ∣) + x

Add a constant to the solution

= 4 (\frac{5}{8} ln 4 x^{2} - 20 x + 24 - \frac{13}{8} ln ∣ 2 x - 4 ∣ + \frac{13}{8} ln ∣ 2 x - 6 ∣) + x + C

What is the integral of (x^2)/(x^2-5x+6) ?
The integral of (x^2)/(x^2-5x+6) is 4(5/8 ln|4x^2-20x+24|-13/8 ln|2x-4|+13/8 ln|2x-6|)+x+C