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

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

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

\int \frac{x ^{4} - 3 x ^{2} - 3 x - 2}{x ^{3} - x ^{2} - 2 x} d x

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

Solution steps

\int \frac{x ^{4} - 3 x ^{2} - 3 x - 2}{x ^{3} - x ^{2} - 2 x} d x

Take the partial fraction of

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

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

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