What is the integral of (x^3+4x-2)/((x-2)(x+1)) ?

The integral of (x^3+4x-2)/((x-2)(x+1)) is (x^2)/2+x+7(1/3 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

integral of (x^3+4x-2)/((x-2)(x+1))

\int \frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} d x

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

Solution steps

\int \frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} d x

Long division

\frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} : x + 1 + \frac{7 x}{x ^{2} - x - 2}

= \int x + 1 + \frac{7 x}{x ^{2} - 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 x d x + \int 1 d x + \int \frac{7 x}{x ^{2} - x - 2} d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

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

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

Add a constant to the solution

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

What is the integral of (x^3+4x-2)/((x-2)(x+1)) ?
The integral of (x^3+4x-2)/((x-2)(x+1)) is (x^2)/2+x+7(1/3 ln|x+1|+2/3 ln|x-2|)+C