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

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

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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