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

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

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

- \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{2}{3 ( x - 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{1}{6 ( x + 1 )} d x - \int \frac{1}{2 ( x - 1 )} d x + \int \frac{2}{3 ( x - 2 )} d x

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

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

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

= - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x - 2 ∣ + C

(4 t^{4} + \frac{4}{t ^{4}})^{\frac{1}{4}}

(e^{2 x} + 4) \frac{d y}{d x} = y

f (x) = (cos (x))^{33} x

\frac{d}{d x} (sin (x) cos (y))

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

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