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

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

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integral of 4/(x^4-1)

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

- 2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 4 \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 )}

= 4 \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

= 4 (- \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 ∣

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

Simplify

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

= - 2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - 2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

\frac{d}{d x} (6 x^{4} + 3)

y = x^{3} - 4 x, y = 12 x

x \to - \infty lim (- 4 e^{x})

y = ln (x^{2} - 3 x + 1), (3, 0)

y^{^{'}} = \frac{x - y sin ( 2 x )}{sin ^{2} ( x )}

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