What is the integral of (x-4)/(x^2-5x+6) ?

The integral of (x-4)/(x^2-5x+6) is -ln|x-3|+2ln|x-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-4)/(x^2-5x+6)

\int \frac{x - 4}{x ^{2} - 5 x + 6} d x

- ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 4}{x ^{2} - 5 x + 6} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 3 u + 2} d u

= \int \frac{u}{u ^{2} + 3 u + 2} d u

Take the partial fraction of

\frac{u}{u ^{2} + 3 u + 2} : - \frac{1}{u + 1} + \frac{2}{u + 2}

= \int - \frac{1}{u + 1} + \frac{2}{u + 2} d u

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}{u + 1} d u + \int \frac{2}{u + 2} d u

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{2}{u + 2} d u = 2 ln ∣ u + 2 ∣

= - ln ∣ u + 1 ∣ + 2 ln ∣ u + 2 ∣

Substitute back

u = x - 4

= - ln ∣ x - 4 + 1 ∣ + 2 ln ∣ x - 4 + 2 ∣

Simplify

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

x \to - 2 lim (\frac{5}{x ^{2}})

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

x \to 1 - lim (\frac{x + 1}{x ^{4} - 1})

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0, y (1) = 0, y^{^{'}} (1) = - 5

\frac{d}{d x} (x^{4} - \frac{x ^{8}}{6} - x^{2} sin (x^{2}))

What is the integral of (x-4)/(x^2-5x+6) ?
The integral of (x-4)/(x^2-5x+6) is -ln|x-3|+2ln|x-2|+C