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

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

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integral of 1/((x^3-x))

integral

\frac{1}{( x ^{3} - x )}

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

Solution steps

\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 )}

= \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

= - \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 ∣

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

Add a constant to the solution

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

7^{2} x + 3

4 x - 1

\frac{( 1 + x )}{x ^{2}}

20 - 2 x

\frac{1}{2 x ^{2}}

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