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

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:

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

-
$\int\big(\frac a x + \frac b {x+1} + \frac c {x-1} +d\big)$ –  Jan Dvorak Jan 17 '13 at 13:33
Try long division –  anorton Jan 17 '13 at 13:49

try this $$\frac{x^3-2}{x^3-x} = 1 + \frac{x-2}{x^3-x}$$ Now use partial fraction.
@Giesi what is $(x^3-2)-(x^3-x)$? –  Jan Dvorak Jan 17 '13 at 13:38