<s>[INST] 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:
$$ \int\frac{x^3-2}{x^3-x} = \frac{x^3-2}{x(x-1)(x+1)} = \frac{x^3}{x^3-x} - \frac{2}{x^3-x}$$
Do you have some advice?
 [/INST] try this 
$$\frac{x^3-2}{x^3-x} = 1  + \frac{x-2}{x^3-x}$$
Now use partial fraction.
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