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cherrilyn

  • 5 years ago

evaluate the integral

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  1. cherrilyn
    • 5 years ago
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    \[\int\limits_{?}^{?}100xdx/(x-3)(x ^{2}+1)^{2}\] I know I change it to A/x-1 + B(x-1)^2+C(x-1)^3 then multiply it by (x-1)^3 then what

  2. anonymous
    • 5 years ago
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    are you sure?

  3. cherrilyn
    • 5 years ago
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    oops I mean A/x-3 + Bx+C/x^2+1 + Dx+E/(x^2+1)^2

  4. anonymous
    • 5 years ago
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    lol

  5. anonymous
    • 5 years ago
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    alright, good , now start taking small values for x ^_^ to find A, B, C and D

  6. anonymous
    • 5 years ago
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    wait, first you have to multiply the denominator on all parts

  7. anonymous
    • 5 years ago
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    Take x=3. You will find A directly.

  8. anonymous
    • 5 years ago
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    A=3

  9. anonymous
    • 5 years ago
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    so, you'll get :\[100x = A(x+1)^2 + Bx + C(x-3)(x^2+1) + Dx + E (x-3)\]

  10. cherrilyn
    • 5 years ago
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    ok A=3 :)

  11. cherrilyn
    • 5 years ago
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    wait no

  12. anonymous
    • 5 years ago
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    now you can take values for x to find all the alphabets ^_^

  13. anonymous
    • 5 years ago
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    hmm, what?

  14. cherrilyn
    • 5 years ago
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    when you distribute in..for example.. Bx+C(x-3)(x^2+1) ... do you distribute the two things in the parenthesis to the Bx AND the C? or just the C

  15. cherrilyn
    • 5 years ago
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    and Distribute (x-3) to the Dx AND the E?

  16. anonymous
    • 5 years ago
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    hmm, I think it's the whole thing

  17. anonymous
    • 5 years ago
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    yes yes, I'm sure

  18. cherrilyn
    • 5 years ago
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    okay so A is indeed 3

  19. anonymous
    • 5 years ago
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    yes it is ^_^

  20. anonymous
    • 5 years ago
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    The better way is to open all brackets, and add like terms together.

  21. anonymous
    • 5 years ago
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    now take other values for x, such as -1 1 2 -2 ~

  22. anonymous
    • 5 years ago
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    B=0

  23. anonymous
    • 5 years ago
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    Oh wait!!

  24. cherrilyn
    • 5 years ago
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    blaah do I have to test all of those numbers? lol its kiind of hard to predict the right values just by looking at it

  25. anonymous
    • 5 years ago
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    yes it's a must, you need to get all the values, A, B,C,D,and E

  26. cherrilyn
    • 5 years ago
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    I mean the numbers -1, 1, 2, and -2

  27. anonymous
    • 5 years ago
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    yes , you must lol, what other way is there to solve it?

  28. anonymous
    • 5 years ago
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    you must check with all values my dear :)

  29. anonymous
    • 5 years ago
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    B=-3

  30. anonymous
    • 5 years ago
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    C=9

  31. anonymous
    • 5 years ago
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    you checked the values? anwar? or shall I check after you lol

  32. anonymous
    • 5 years ago
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    D=18

  33. cherrilyn
    • 5 years ago
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    Okay If I plug in A=3 and x=-1..... I get -100=12+2B-2C+4D-4E. now what? Or am I doing this wrong

  34. anonymous
    • 5 years ago
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    no you're write, so now you have equation 1, you can use it later to either subtract or add from another equation ^_^

  35. anonymous
    • 5 years ago
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    right*

  36. anonymous
    • 5 years ago
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    alright I gtg, I'll be back later, bye ^_^

  37. cherrilyn
    • 5 years ago
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    okay thankss! Ttyl :)

  38. anonymous
    • 5 years ago
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    anwar will continue, most welcome :)

  39. anonymous
    • 5 years ago
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    I got all of them. Hopefully right!

  40. anonymous
    • 5 years ago
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    A=3, B=-3, C=9, D=24, E=-8

  41. anonymous
    • 5 years ago
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    Bye starica!!

  42. cherrilyn
    • 5 years ago
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    thanks anwar!

  43. anonymous
    • 5 years ago
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    You're welcome!! You may want to check them.

  44. anonymous
    • 5 years ago
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    OMG, Sorry there was a mistake.

  45. anonymous
    • 5 years ago
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    It should be: A=3, B=-3, C=-9, D=-30, E=10. Sorry for the confusion.

  46. cherrilyn
    • 5 years ago
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    what values for x did you use?

  47. anonymous
    • 5 years ago
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    Well.. I opened all the parenthesis, and add like terms together, and then solved the equations.

  48. anonymous
    • 5 years ago
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    For example if you have x^2(A+B)+x(C+2B)=5x^2+x, that means: A+BC=5 and C+2B=1.. I used this idea.

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