Question

How do you integrate `(x^2+2x-1)/(x^3-x)dx`?

Answer 1

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

Explanation:

`(x^2+2x-1)/(x^3-x) = (x^2+2x-1)/(x(x-1)(x+1)) = a/x + b/(x-1) + c/(x+1)`

Use Heaviside's cover up method to find `a, b, c`:

`a = ((color(blue)(0))^2+2(color(blue)(0))-1)/(((color(blue)(0))-1)((color(blue)(0))+1)) = (-1)/((-1)(1)) = 1`

`b = ((color(blue)(1))^2+2(color(blue)(1))-1)/((color(blue)(1))((color(blue)(1))+1)) = (2)/((1)(2)) = 1`

`c = ((color(blue)(-1))^2+2(color(blue)(-1))-1)/((color(blue)(-1))((color(blue)(-1))-1)) = (-2)/((-1)(-2)) = -1`

So:

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

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