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anonymous

  • 5 years ago

i need to find the antiderivative of 1/(x+4)^3 using the method of partial fractions

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  1. anonymous
    • 5 years ago
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    y partial fractions when u can find the antiderivative directly?

  2. amistre64
    • 5 years ago
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    partial are cooler lol

  3. anonymous
    • 5 years ago
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    lol i said the same thing but that is the unit we are on!

  4. anonymous
    • 5 years ago
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    :)

  5. amistre64
    • 5 years ago
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    A + B + C ----- ------- -------- (x+4) (x+4)^2 (x+4)^3

  6. anonymous
    • 5 years ago
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    correct now solve it and u get A= B= 0 :P

  7. amistre64
    • 5 years ago
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    A(x+4)^2 ------------ (x+4)(x+4)^2

  8. amistre64
    • 5 years ago
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    A(x+4)^2 +B(x+4)+ C 1 ---------------------- = ------- (x+4)^3 (x+4)^3 right?

  9. anonymous
    • 5 years ago
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    i thought you had to do it where you cancel denominators and set 1= A(x+4)2 + B(x+4) +C and then solve

  10. anonymous
    • 5 years ago
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    *A(x+4)^2

  11. amistre64
    • 5 years ago
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    you tend to split them to see what you had to multiply by to get common denominators

  12. amistre64
    • 5 years ago
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    then the tops equate

  13. anonymous
    • 5 years ago
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    ohh okay great!

  14. amistre64
    • 5 years ago
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    A(x+4)^2 +B(x+4)+ C = 1 A(x^2 +8x +16) +Bx +B4 +C = 1

  15. amistre64
    • 5 years ago
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    Ax^2 +A8x +A16 +Bx +B4 +C = 1

  16. amistre64
    • 5 years ago
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    this looking alright so far?

  17. amistre64
    • 5 years ago
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    x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

  18. anonymous
    • 5 years ago
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    im working on it :)

  19. amistre64
    • 5 years ago
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    forgot the B4... 16A +4B +C = 1

  20. amistre64
    • 5 years ago
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    i get to here and draw a blank lol

  21. amistre64
    • 5 years ago
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    do we find a value for a variable and stick it in?

  22. anonymous
    • 5 years ago
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    yes we need to be able to know what A B and C are equal to

  23. amistre64
    • 5 years ago
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    a = 0

  24. amistre64
    • 5 years ago
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    x^2A = x^2(0) A = 0 right?

  25. amistre64
    • 5 years ago
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    16(0) +4(0) +C = 1 C = 1

  26. amistre64
    • 5 years ago
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    we equate equal parts in this right? x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

  27. amistre64
    • 5 years ago
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    (0)x^2 = Ax^2; A = 0 (0)x = (8A+B)x = Bx; B=0 A16 +4B + C = 1 0+0+C = 1 c = 1

  28. anonymous
    • 5 years ago
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    yeah

  29. amistre64
    • 5 years ago
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    so the derivative is simply the derivative of the original equation :)

  30. anonymous
    • 5 years ago
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    ugh oh. let me get back to you i think we may have gone the wrong rout earlier would you like to work through an easier problem so i can show you the jist of it and then maybe we can come back to solving this one? feel free to say no

  31. anonymous
    • 5 years ago
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    nvm u are right and the antiderivative is the same as the orginal solve the shorter way

  32. amistre64
    • 5 years ago
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    im game .....

  33. amistre64
    • 5 years ago
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    if they had given a factored form of an equation inthe bottom with different fillers, it would have been way more exciting lol

  34. anonymous
    • 5 years ago
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    yeah like 1/(x-1)(x+2)^2

  35. anonymous
    • 5 years ago
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    lamee

  36. amistre64
    • 5 years ago
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    yeah, that would ve been awesome!! lol

  37. anonymous
    • 5 years ago
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    so i see you enjoy math lol? im a soph taking calc II

  38. anonymous
    • 5 years ago
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    thaks for your help by the way

  39. amistre64
    • 5 years ago
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    and a polynomial up top to liven things up ;)

  40. amistre64
    • 5 years ago
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    ill be takinng calc 1 over the summer in community college

  41. amistre64
    • 5 years ago
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    youre welcome :)

  42. anonymous
    • 5 years ago
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    cool cool good for you!

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