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

The integral of x/(x^3-3x+2) is 2/9 ln|x-1|-1/(3(x-1))-2/9 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^3-3x+2)

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

\frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{x ^{3} - 3 x + 2} : \frac{2}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} - \frac{2}{9 ( x + 2 )}

= \int \frac{2}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} - \frac{2}{9 ( 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{2}{9 ( x - 1 )} d x + \int \frac{1}{3 ( x - 1 ) ^{2}} d x - \int \frac{2}{9 ( x + 2 )} d x

\int \frac{2}{9 ( x - 1 )} d x = \frac{2}{9} ln ∣ x - 1 ∣

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

\int \frac{2}{9 ( x + 2 )} d x = \frac{2}{9} ln ∣ x + 2 ∣

= \frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} ln ∣ x + 2 ∣ + C

\int 5^{x} cos (x) d x

\int tan (- x) d x

\int \frac{16 x ^{2} - 1}{x ^{3}} d x

\int 2 e^{x} + x^{2} d x

\int \frac{1}{( 2 + sin ( x ) )} d x

What is the integral of x/(x^3-3x+2) ?
The integral of x/(x^3-3x+2) is 2/9 ln|x-1|-1/(3(x-1))-2/9 ln|x+2|+C