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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{3} - 1}{x ^{2} - 1} : \frac{x ^{3}}{x ^{2} - 1} - \frac{1}{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{1}{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{1}{x ^{2} - 1} d x = - \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 ∣)

Simplify

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

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

Add a constant to the solution

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

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