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inkyvoyd

  • 2 years ago

Integrate \(\Huge \frac{x}{ax^2+bx+c}\)

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  1. waterineyes
    • 2 years ago
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    Linear by Quadratic...

  2. inkyvoyd
    • 2 years ago
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    Eh?

  3. inkyvoyd
    • 2 years ago
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    But, how exactly am I supposed to do this one?

  4. waterineyes
    • 2 years ago
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    If you put ax2 + bx + c = y then it helps or not??

  5. inkyvoyd
    • 2 years ago
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    uhh... what about the top - we just substitute using the quadratic formula in that case?

  6. inkyvoyd
    • 2 years ago
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    Sorry guys, I'm not even sure where to start with this integral. I can't find it in my book either >.<

  7. eyust707
    • 2 years ago
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    bring everything on the bottom to the top... then integration by parts?

  8. inkyvoyd
    • 2 years ago
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    uhh - I'm not sure that will work >.<

  9. eyust707
    • 2 years ago
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    why not...

  10. inkyvoyd
    • 2 years ago
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    All the related integrals had something to do with trig sub and inverse trig functions, but i can't find this case.

  11. eyust707
    • 2 years ago
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    ohh

  12. inkyvoyd
    • 2 years ago
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    I know what the answer is (from an integral table), and it's logarithmic+trigonometric. I have no idea where to start though

  13. inkyvoyd
    • 2 years ago
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    http://integral-table.com/downloads/single-page-integral-table.pdf #16, btw

  14. eyust707
    • 2 years ago
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    ahhh that got me thinkin... let me take another look

  15. eyust707
    • 2 years ago
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    well log came from 1/x

  16. inkyvoyd
    • 2 years ago
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    Yes.

  17. eyust707
    • 2 years ago
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    that looks like a big u sub

  18. inkyvoyd
    • 2 years ago
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    Uh, got any ideas where to start htough?

  19. eyust707
    • 2 years ago
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    one sec

  20. inkyvoyd
    • 2 years ago
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    kay

  21. eyust707
    • 2 years ago
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    just by the looks w/o really thinking about it, it looks like they let the bottom = u

  22. eyust707
    • 2 years ago
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    then factor out a 2a

  23. eyust707
    • 2 years ago
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    1/2a *

  24. eyust707
    • 2 years ago
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    wait wtf

  25. eyust707
    • 2 years ago
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    lol nvm i thought the second half was the next step

  26. eyust707
    • 2 years ago
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    thats a tricky one i need a piece of paper for that one lol

  27. inkyvoyd
    • 2 years ago
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    okay lol. Thanks for helping :P

  28. eyust707
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=int+x%2F+%28ax2+%2B+bx+%2B+c%29

  29. eyust707
    • 2 years ago
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    that way is crazzyyyy tho lol

  30. inkyvoyd
    • 2 years ago
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    lol - but the answer is different from the integral table :S the result is too complicated lol. I'm using some constants for the integral that will make it even more nasty x.x

  31. eyust707
    • 2 years ago
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    its the same just written different

  32. experimentX
    • 2 years ago
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    x = 1/2(2x + b/a) - 1/2 (b/a) make two fractions ... and use the usual approach

  33. eyust707
    • 2 years ago
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    ^^^ thats what w.a did too

  34. inkyvoyd
    • 2 years ago
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    Okay. any way to reduce it to the form seen in the integral table?

  35. eyust707
    • 2 years ago
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    probably not

  36. inkyvoyd
    • 2 years ago
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    once i substitute my values in, it is still an intermediate result - I still have to do two more substitutions, and I'm not sure I can do the algebra without going insane.

  37. experimentX
    • 2 years ago
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    one is going to be log another ... complete squares (a+sqrt(b/4a))^2 - (b/4a-c)

  38. experimentX
    • 2 years ago
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    this is in standard form ... you should be able to evaluate this.

  39. inkyvoyd
    • 2 years ago
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    Wait, I can do it by completing the square as well?

  40. experimentX
    • 2 years ago
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    \[ \frac1{2a}\frac{x + b/a}{x + b/a x+c/a} - \frac1{2a}\frac{ b/a}{(x + \sqrt{b/4a})^2 - (b/4a-c/a)}\]

  41. inkyvoyd
    • 2 years ago
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    oh wow. this integral - sqrt(tan x) - is a freaking pain...

  42. experimentX
    • 2 years ago
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    quite a bit ... assuming c/a > b/4a

  43. eyust707
    • 2 years ago
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    yea just about all realistic functions are either extremely tedious or impossible to solve by hand. When it comes to practical application, we usually just have computers solve or approximate the solution.

  44. inkyvoyd
    • 2 years ago
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    Ugh, I give up. I'm tired of this integral lol. I'll look into it more tomorrow. Thanks though, @experimentX and @eyust707

  45. eyust707
    • 2 years ago
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    thats when the mathematicians come in handy =P

  46. eyust707
    • 2 years ago
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    yw inky.. you taking calc 2 over the summer??

  47. inkyvoyd
    • 2 years ago
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    I'm trying to finish calc 3 and linear algebra and diff equations over the summer >.<

  48. eyust707
    • 2 years ago
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    crazyyy

  49. eyust707
    • 2 years ago
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    that is a lot of material to learn!!

  50. inkyvoyd
    • 2 years ago
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    Yea, hopefully I can finish it.

  51. eyust707
    • 2 years ago
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    I got faith!! where about are you in diffs?

  52. inkyvoyd
    • 2 years ago
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    Haven't even started LOL

  53. eyust707
    • 2 years ago
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    haha I see I see

  54. inkyvoyd
    • 2 years ago
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    well, I skimmed the variables seperatble and the first order linear ODE I think

  55. experimentX
    • 2 years ago
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    It's just a substitution ... reduces to the form \[ \int \frac1 {x^2 + a^2}dx\] or, \[ \int \frac 1{x^2 - a^2}dx\] assuming first, wolf calculates 2/a arctan(x/a)

  56. eyust707
    • 2 years ago
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    yea seps will seem very natural

  57. eyust707
    • 2 years ago
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    linear first order with the integrating factor is tricky to understand at first... but easy to do

  58. eyust707
    • 2 years ago
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    youll prob see frustrating word problems but dont let them scare you.. they always work out to be quite simple

  59. inkyvoyd
    • 2 years ago
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    alright - I have to learn how to integrate these nasty partial fractions first though haha :P

  60. eyust707
    • 2 years ago
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    haha just remember, If your integrating with respect to y , all the other variables are constant!

  61. eyust707
    • 2 years ago
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    keep repeating that in your head as you run through your derivative..

  62. eyust707
    • 2 years ago
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    pretend that all the other letters are just numbers and youll be golden!

  63. dpaInc
    • 2 years ago
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    didn't you see what i last posted earlier inky?

  64. inkyvoyd
    • 2 years ago
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    uhh what did you post?

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