Question

How do you integrate `int 1/(x^3 - x)` using partial fractions?

Answer 1

`int 1/(x^3-x) dx = -ln abs(x)+1/2 ln abs(x-1)+1/2 ln abs(x+1) + C`

Explanation:

`x^3-x = x(x-1)(x+1)`

So:

`1/(x^3-x) = 1/(x(x-1)(x+1)) = A/x+B/(x-1)+C/(x+1)`

Using Heaviside's cover up method we find:

`A=1/(((color(blue)(0))-1)((color(blue)(0))+1)) = 1/((-1)(1)) = -1`

`B=1/((color(blue)(1))((color(blue)(1))+1))=1/((1)(2))=1/2`

`C=1/((color(blue)(-1))((color(blue)(-1))-1)) = 1/((-1)(-2))=1/2`

So:

`int 1/(x^3-x) dx = int (-1/x+1/(2(x-1))+1/(2(x+1))) dx`

`color(white)(int 1/(x^3-x) dx) = -ln abs(x)+1/2 ln abs(x-1)+1/2 ln abs(x+1) + C`