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

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