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

The integral of (x^4)/((x^2-1)(x-2)) is (x^2)/2+2x+1/6 ln|x+1|-1/2 ln|x-1|+16/3 ln|x-2|+C

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

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

\frac{x ^{2}}{2} + 2 x + \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{( x ^{2} - 1 ) ( x - 2 )} : x + 2 + \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{16}{3 ( x - 2 )}

= \int x + 2 + \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{16}{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 2 d x + \int \frac{1}{6 ( x + 1 )} d x - \int \frac{1}{2 ( x - 1 )} d x + \int \frac{16}{3 ( x - 2 )} d x

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

\int 2 d x = 2 x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

\int \frac{16}{3 ( x - 2 )} d x = \frac{16}{3} ln ∣ x - 2 ∣

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

Add a constant to the solution

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

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What is the integral of (x^4)/((x^2-1)(x-2)) ?
The integral of (x^4)/((x^2-1)(x-2)) is (x^2)/2+2x+1/6 ln|x+1|-1/2 ln|x-1|+16/3 ln|x-2|+C