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

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

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integral of 3/(x^3-x)

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} - x} : - \frac{1}{x} + \frac{1}{2 ( x + 1 )} + \frac{1}{2 ( x - 1 )}

= 3 \cdot \int - \frac{1}{x} + \frac{1}{2 ( x + 1 )} + \frac{1}{2 ( 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}{x} d x + \int \frac{1}{2 ( x + 1 )} d x + \int \frac{1}{2 ( x - 1 )} d x)

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

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