What is the integral of (x^3+1)/(x^3-3x+2) ?

The integral of (x^3+1)/(x^3-3x+2) is x+7/9 ln|x-1|-7/9 ln|x+2|-1/(x-1)+1/(3(x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3+1)/(x^3-3x+2)

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

x + \frac{7}{9} ln ∣ x - 1 ∣ - \frac{7}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{1}{3 ( x - 1 )} + C

Solution steps

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

Expand

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

\int \frac{x ^{3}}{x ^{3} - 3 x + 2} d x = x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )}

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

= x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )} - \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} + \frac{1}{9} ln ∣ x + 2 ∣

Simplify

x + \frac{8}{9} ln ∣ x - 1 ∣ - \frac{8}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{2}{3 ( x - 1 )} - \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} + \frac{1}{9} ln ∣ x + 2 ∣ : x + \frac{7}{9} ln ∣ x - 1 ∣ - \frac{7}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{1}{3 ( x - 1 )}

= x + \frac{7}{9} ln ∣ x - 1 ∣ - \frac{7}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{1}{3 ( x - 1 )}

Add a constant to the solution

= x + \frac{7}{9} ln ∣ x - 1 ∣ - \frac{7}{9} ln ∣ x + 2 ∣ - \frac{1}{x - 1} + \frac{1}{3 ( x - 1 )} + C

\int \frac{1}{sin ^{2} ( x ) + 5 cos ^{2} ( x )} d x

\int \frac{1}{( 8 + x ) ^{2}} d x

\int \frac{x ^{3}}{4 - x ^{8}} d x

\int (2 x^{2} + 1) ln (x) d x

\int 3 arctan (3 x) d x

What is the integral of (x^3+1)/(x^3-3x+2) ?
The integral of (x^3+1)/(x^3-3x+2) is x+7/9 ln|x-1|-7/9 ln|x+2|-1/(x-1)+1/(3(x-1))+C