Integral of $\frac{x^3-2}{x^3-x}$

Question

As the title says, I need some help with this function: $$ \int\frac{x^3-2}{x^3-x}$$ I tried it with different versions/forms of the function to get it with substitution or partiel integration:

$$ \int\frac{x^3-2}{x^3-x} = \frac{x^3-2}{x(x-1)(x+1)} = \frac{x^3}{x^3-x} - \frac{2}{x^3-x}$$

Do you have some advice?

Answer 1

try this $$\frac{x^3-2}{x^3-x} = 1 + \frac{x-2}{x^3-x}$$ Now use partial fraction.