## baldymcgee6 How to setup this integral with partial fractions..? one year ago one year ago

1. baldymcgee6

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2. baldymcgee6

x^7

3. baldymcgee6

just have to set it up, don't have to actually integrate it. :)

4. hartnn

x(x^6-1) =x(x^3+1)(x^3-1)

5. baldymcgee6

$\LARGE{\int\limits \frac{ 1 }{ x^7 - x }dx}$

6. kirbykirby

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7. hartnn

do you know general formulas for a^3+b^3 and a^3-b^3..

8. baldymcgee6

yes

9. baldymcgee6

so how would this spit into partials?

10. hartnn

if there is quadratic equation in denominator, which cannot be factorized, then in numerator, it'll be linear expression, Ax+B...

11. kirbykirby

For x^3-1 and x^3+1 you can use Long polynomial division by finding a factor (x-a) where a is a root of the polynomial

12. hartnn

so, whats x^3+1=... ? x^3-1=.... ?

13. kirbykirby

I never learnt the general formula for a^3+b^3. I think it's easier to do polynomial division because it will work for higher powers

14. baldymcgee6

x^3 +1 = (x+1)(x^2-x+1)

15. hartnn

so, above x^2-x+1 will come a linear expression like Ax+B and above linear terms like x, x(x+1), constants like C,D,E ... will come.

16. baldymcgee6

@kirbykirby polynomial division of what and what?

17. baldymcgee6

x^3 -1 = (x-1)(x^2+x+1)

18. baldymcgee6

okay, i'll try this

19. hartnn

so, you have 5 factors, out of which 2 is quadratic.

20. kirbykirby

for x^3+1, you find the factors of the constant (here, its 1) and find which one = 0 when you plug it into your x-terms. So, -1 works: ((-1)^3+1) = 0. So, the factor that divides the polynomial is always of the form (x-a). So, (x-(-1)) = (x+1). Then divide x^3+1 into x+1

21. baldymcgee6

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22. hartnn

$$\checkmark$$

23. baldymcgee6

@kirbykirby that is useful to know rather than memorizing these formulas.

24. baldymcgee6

thanks for the help guys!

25. kirbykirby

No problem :)

26. kirbykirby

It's often called the "Factor Theorem". http://en.wikipedia.org/wiki/Factor_theorem