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

The integral of (x^4)/(x^2-x-2) is (x^3)/3+(x^2)/2+3x-1/3 ln|x+1|+16/3 ln|x-2|+C

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

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

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 3 x - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

Solution steps

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

Long division

\frac{x ^{4}}{x ^{2} - x - 2} : x^{2} + x + 3 + \frac{5 x + 6}{x ^{2} - x - 2}

= \int x^{2} + x + 3 + \frac{5 x + 6}{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^{2} d x + \int x d x + \int 3 d x + \int \frac{5 x + 6}{x ^{2} - x - 2} d x

\int x^{2} d x = \frac{x ^{3}}{3}

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

\int 3 d x = 3 x

\int \frac{5 x + 6}{x ^{2} - x - 2} d x = - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 3 x - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 3 x - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

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What is the integral of (x^4)/(x^2-x-2) ?
The integral of (x^4)/(x^2-x-2) is (x^3)/3+(x^2)/2+3x-1/3 ln|x+1|+16/3 ln|x-2|+C