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

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

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integral of 1/(x^4-1)

\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 ∣ + C

Solution steps

\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 )}

= \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

= - \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 ∣

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

Add a constant to the solution

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

x \to 2 lim (\frac{2 - x}{4 - x ^{2}})

\frac{d}{d x} (x^{3} + 2 x y + y^{2} + z^{2} - 5 x - 4 z)

x \to 9 lim (\frac{81}{9 - x})

x^{2} + \frac{4}{x} - 10

3 x - 4 x^{2}

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