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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?

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

1 Answer 1

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

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Why is it the same? –  Giesi Jan 17 '13 at 13:36
1  
take LCM and see –  Santosh Linkha Jan 17 '13 at 13:37
2  
@Giesi what is $(x^3-2)-(x^3-x)$? –  Jan Dvorak Jan 17 '13 at 13:38
    
Thx I get it ;) –  Giesi Jan 17 '13 at 13:43

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