What is the integral of (x^9-3x+1)/(x^2-1) ?

The integral of (x^9-3x+1)/(x^2-1) is

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integral of (x^9-3x+1)/(x^2-1)

\int \frac{x ^{9} - 3 x + 1}{x ^{2} - 1} d x

\frac{x ^{8}}{8} + \frac{x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - ln x^{2} - 1 - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{9} - 3 x + 1}{x ^{2} - 1} d x

Long division

\frac{x ^{9} - 3 x + 1}{x ^{2} - 1} : x^{7} + x^{5} + x^{3} + x + \frac{- 2 x + 1}{x ^{2} - 1}

= \int x^{7} + x^{5} + x^{3} + x + \frac{- 2 x + 1}{x ^{2} - 1} 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^{7} d x + \int x^{5} d x + \int x^{3} d x + \int x d x + \int \frac{- 2 x + 1}{x ^{2} - 1} d x

\int x^{7} d x = \frac{x ^{8}}{8}

\int x^{5} d x = \frac{x ^{6}}{6}

\int x^{3} d x = \frac{x ^{4}}{4}

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

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

= \frac{x ^{8}}{8} + \frac{x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - ln x^{2} - 1 - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - ln x^{2} - 1 - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

What is the integral of (x^9-3x+1)/(x^2-1) ?
The integral of (x^9-3x+1)/(x^2-1) is