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

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

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

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

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

Add a constant to the solution

= x - \frac{1}{2} arctan (x) - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{4} 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^4+x^2-x+1)/(x^4-1) ?
The integral of (x^4+x^2-x+1)/(x^4-1) is