What is the integral of (x^3-x^2-x-3)/(x^4-1) ?

The integral of (x^3-x^2-x-3)/(x^4-1) is

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integral of (x^3-x^2-x-3)/(x^4-1)

\int \frac{x ^{3} - x ^{2} - x - 3}{x ^{4} - 1} d x

\frac{1}{4} ln x^{4} - 1 + arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} - 1 + C

Solution steps

\int \frac{x ^{3} - x ^{2} - x - 3}{x ^{4} - 1} d x

Expand

\frac{x ^{3} - x ^{2} - x - 3}{x ^{4} - 1} : \frac{x ^{3}}{x ^{4} - 1} - \frac{x ^{2}}{x ^{4} - 1} - \frac{x}{x ^{4} - 1} - \frac{3}{x ^{4} - 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 ^{4} - 1} d x - \int \frac{x ^{2}}{x ^{4} - 1} d x - \int \frac{x}{x ^{4} - 1} d x - \int \frac{3}{x ^{4} - 1} d x

\int \frac{x ^{3}}{x ^{4} - 1} d x = \frac{1}{4} ln x^{4} - 1

\int \frac{x ^{2}}{x ^{4} - 1} d x = \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

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

\int \frac{3}{x ^{4} - 1} d x = 3 (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

= \frac{1}{4} ln x^{4} - 1 - (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) - (- \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)) - 3 (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

Simplify

\frac{1}{4} ln x^{4} - 1 - (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) - (- \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)) - 3 (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) : \frac{1}{4} ln x^{4} - 1 + arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} - 1

= \frac{1}{4} ln x^{4} - 1 + arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} - 1

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 1 + arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{4} ln x^{2} - 1 + C

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What is the integral of (x^3-x^2-x-3)/(x^4-1) ?
The integral of (x^3-x^2-x-3)/(x^4-1) is