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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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