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

The integral of (2x^4+4x^3-2)/(x^4-1) is 2x+ln|x^4-1|+C

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

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

2 x + ln x^{4} - 1 + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= 2 x + ln x^{4} - 1

Add a constant to the solution

= 2 x + ln x^{4} - 1 + C

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