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baldymcgee6 Group Title

How to setup this integral with partial fractions..?

  • one year ago
  • one year ago

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  1. baldymcgee6 Group Title
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    |dw:1358744328225:dw|

    • one year ago
  2. baldymcgee6 Group Title
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    x^7

    • one year ago
  3. baldymcgee6 Group Title
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    just have to set it up, don't have to actually integrate it. :)

    • one year ago
  4. hartnn Group Title
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    x(x^6-1) =x(x^3+1)(x^3-1)

    • one year ago
  5. baldymcgee6 Group Title
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    \[\LARGE{\int\limits \frac{ 1 }{ x^7 - x }dx}\]

    • one year ago
  6. kirbykirby Group Title
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    |dw:1358744642346:dw|

    • one year ago
  7. hartnn Group Title
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    do you know general formulas for a^3+b^3 and a^3-b^3..

    • one year ago
  8. baldymcgee6 Group Title
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    yes

    • one year ago
  9. baldymcgee6 Group Title
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    so how would this spit into partials?

    • one year ago
  10. hartnn Group Title
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    if there is quadratic equation in denominator, which cannot be factorized, then in numerator, it'll be linear expression, Ax+B...

    • one year ago
  11. kirbykirby Group Title
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    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

    • one year ago
  12. hartnn Group Title
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    so, whats x^3+1=... ? x^3-1=.... ?

    • one year ago
  13. kirbykirby Group Title
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    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

    • one year ago
  14. baldymcgee6 Group Title
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    x^3 +1 = (x+1)(x^2-x+1)

    • one year ago
  15. hartnn Group Title
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    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.

    • one year ago
  16. baldymcgee6 Group Title
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    @kirbykirby polynomial division of what and what?

    • one year ago
  17. baldymcgee6 Group Title
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    x^3 -1 = (x-1)(x^2+x+1)

    • one year ago
  18. baldymcgee6 Group Title
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    okay, i'll try this

    • one year ago
  19. hartnn Group Title
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    so, you have 5 factors, out of which 2 is quadratic.

    • one year ago
  20. kirbykirby Group Title
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    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

    • one year ago
  21. baldymcgee6 Group Title
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    |dw:1358745155474:dw|

    • one year ago
  22. hartnn Group Title
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    \(\checkmark \)

    • one year ago
  23. baldymcgee6 Group Title
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    @kirbykirby that is useful to know rather than memorizing these formulas.

    • one year ago
  24. baldymcgee6 Group Title
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    thanks for the help guys!

    • one year ago
  25. kirbykirby Group Title
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    No problem :)

    • one year ago
  26. kirbykirby Group Title
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    It's often called the "Factor Theorem". http://en.wikipedia.org/wiki/Factor_theorem

    • one year ago
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