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

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

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integral of 2/(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 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} arctan (x)

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

x \to 4 lim (9)

x \to 0 lim (v ec t or)

a re a 2 x^{2} + 4 x - 2, 0, 2

f (x) = (5 x^{3} + 2)^{4}

\frac{d}{d x} (tan (arcsin (x)))

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