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

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

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integral of 3/(x^4-1)

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

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

Solution steps

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

Take the constant out:

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

= 3 \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 )}

= 3 \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

= 3 (- \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 ∣

= 3 (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

Add a constant to the solution

= 3 (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + C

sin (x) tan (x)

\frac{d}{d x} (3 cos (5 x) + sin (5 x))

\int_{1}^{16} \frac{1}{2 x} d x

\int \frac{- x + 8}{- x ^{2} - x + 8} d x

x \to 1 lim (\frac{1 + x}{1 - x})

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