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cutelilgirl Group Title

how to integrate log | 4x^2-1| dx ?

  • 2 months ago
  • 2 months ago

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  1. satellite73 Group Title
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    start by factoring

    • 2 months ago
  2. satellite73 Group Title
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    then break apart in to two pieces then integrate term by term

    • 2 months ago
  3. Elsa213 Group Title
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    erm.... integral 4x/(1+x^2) = I Let 1+x^2 = t ...........(1) --> 2x = dt/dx --> 2x dx = dt ...........(2) Now I = integral 2 [2x dx / (1+x^2) ] = integral 2 [dt / t ] since from (2) = 2 [integral dt / t ] = 2 (ln t) +c = 2 ln(1+x^2) + c We have integral 1/t dt = ln t +c Therefore I = 2 ln(1+x^2) +c where ln is log to the base e https://answers.yahoo.com/question/index?qid=20110416060345AAVswO7

    • 2 months ago
  4. satellite73 Group Title
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    @Elsa213 i am pretty sure that is some other answer to some other question

    • 2 months ago
  5. Elsa213 Group Title
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    ik ik

    • 2 months ago
  6. Elsa213 Group Title
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    it was just an example

    • 2 months ago
  7. cutelilgirl Group Title
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    log | (2x+1)(2x-1) | ??

    • 2 months ago
  8. xapproachesinfinity Group Title
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    log a.b=loga +logb then integrate each at a time

    • 2 months ago
  9. satellite73 Group Title
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    right

    • 2 months ago
  10. satellite73 Group Title
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    now \[\int \log(2x+1)dx+\int (2x-1)dx\]

    • 2 months ago
  11. xapproachesinfinity Group Title
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    you forgot log lol

    • 2 months ago
  12. satellite73 Group Title
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    then you should be done because you should have the integral of the log memorized

    • 2 months ago
  13. satellite73 Group Title
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    so i did \[\int \log(2x+1)dx+\int \log(2x-1)dx\]

    • 2 months ago
  14. cutelilgirl Group Title
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    the answer given is : x log | 4x^2-1| - 1/2 log | (2x-1)/(2x+1) | .. will i get this answer this way ?

    • 2 months ago
  15. xapproachesinfinity Group Title
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    yes continue this way

    • 2 months ago
  16. cutelilgirl Group Title
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    ok thanks @xapproachesinfinity and @satellite73

    • 2 months ago
  17. xapproachesinfinity Group Title
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    you should know what is int log|x|dx

    • 2 months ago
  18. satellite73 Group Title
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    i have no idea when you integrate you will get something then using the properties of the log you can change it in to many different forms don't get married to the answer in the book

    • 2 months ago
  19. cutelilgirl Group Title
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    because this exercise was about the problems of integral dx / (a^2_x^2) type

    • 2 months ago
  20. satellite73 Group Title
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    i would say this integral looks nothing like that at all

    • 2 months ago
  21. xapproachesinfinity Group Title
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    satellite you assumed that the expression will take positive values only you need the absolute value theere

    • 2 months ago
  22. xapproachesinfinity Group Title
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    i would say this integral looks nothing like that at all === is this to my reply^_^

    • 2 months ago
  23. cutelilgirl Group Title
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    yes the | 4x^2 -1 | was there

    • 2 months ago
  24. satellite73 Group Title
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    i would say \[\int \log(4x^2-1)dx\] looks nothing like \[\int \frac{dx}{a^2-x^2}\]

    • 2 months ago
  25. xapproachesinfinity Group Title
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    yes i doesn't :)

    • 2 months ago
  26. xapproachesinfinity Group Title
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    it*

    • 2 months ago
  27. cutelilgirl Group Title
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    or dx/x^2-a^2

    • 2 months ago
  28. satellite73 Group Title
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    \[\int \log(x)dx=x\log(x)-x\] so \[\int \log(2x+1)dx=\frac{1}{2}\left((2x+1)\log(2x-1)-(2x+1)\right)\] by a mental u sub second part is similar

    • 2 months ago
  29. xapproachesinfinity Group Title
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    @satellite73 , any reason why you took off absolute values?

    • 2 months ago
  30. satellite73 Group Title
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    damn typo \[\int \log(2x+1)dx=\frac{1}{2}\left((2x+1)\log(2x+1)-(2x+1)\right)\]

    • 2 months ago
  31. satellite73 Group Title
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    or if you prefer \[\frac{1}{2}(2x+1)(\log(2x+1)-1)\]

    • 2 months ago
  32. satellite73 Group Title
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    laziness is all that and the stupid \(+ c\) got omitted as well

    • 2 months ago
  33. xapproachesinfinity Group Title
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    but you excluded the negative values?

    • 2 months ago
  34. ganeshie8 Group Title
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    why wolfram is giving this mess http://www.wolframalpha.com/input/?i=%5Cint+log%28abs%284x%5E2-1%29%29

    • 2 months ago
  35. xapproachesinfinity Group Title
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    wolfram is crazy hehe

    • 2 months ago
  36. xapproachesinfinity Group Title
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    i say the absolute value cannot be taken off. as we are excluding negative interval here

    • 2 months ago
  37. satellite73 Group Title
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    wolfram gave that mess because of the sign problem here it is neater http://www.wolframalpha.com/input/?i=\int+log%284x^2-1%29

    • 2 months ago
  38. xapproachesinfinity Group Title
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    \(\Large \rm\color{midnightblue}{\int log|(2x+1)(2x-1)|dx=\int log|2x+1|dx+\\ \int log|2x-1|dx}\)

    • 2 months ago
  39. myininaya Group Title
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    Integration by parts should even work for the first way this expression appears

    • 2 months ago
  40. myininaya Group Title
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    integration by parts will be needed polynomial division partial fractions look like they are needed as well

    • 2 months ago
  41. xapproachesinfinity Group Title
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    i agree. i looks that the answer would look much simpler

    • 2 months ago
  42. xapproachesinfinity Group Title
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    x/(2x-1)(2x+1) partial fraction here

    • 2 months ago
  43. xapproachesinfinity Group Title
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    the other part is good xlog|4x^2-1|

    • 2 months ago
  44. myininaya Group Title
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    Whenever I see something that looks like \[\int\limits_{}^{}1 \cdot \ln(f(x)) dx \] one of my first instincts it to go with the integration by parts (assuming f>0) \[=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx\]

    • 2 months ago
  45. xapproachesinfinity Group Title
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    so far it looks nice to me

    • 2 months ago
  46. myininaya Group Title
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    where did you get that @xapproachesinfinity ?

    • 2 months ago
  47. xapproachesinfinity Group Title
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    |dw:1408559618618:dw|

    • 2 months ago
  48. xapproachesinfinity Group Title
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    from by part integration

    • 2 months ago
  49. xapproachesinfinity Group Title
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    this now going to that expression she wrote as the answer!

    • 2 months ago
  50. myininaya Group Title
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    I don't see where that integral comes from

    • 2 months ago
  51. myininaya Group Title
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    i wonder if you forget f' on top |dw:1408559852906:dw|

    • 2 months ago
  52. xapproachesinfinity Group Title
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    u=log(|4x^2-1|) ===> du=1/4x^2-1 dv=1 ====> v=x may be there is an error lol

    • 2 months ago
  53. myininaya Group Title
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    derivative of ln(f) is equal to f'/f

    • 2 months ago
  54. xapproachesinfinity Group Title
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    yup you are right forgot that xD

    • 2 months ago
  55. xapproachesinfinity Group Title
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    it should be what you wrote above

    • 2 months ago
  56. myininaya Group Title
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    @cutelilgirl are you still there?

    • 2 months ago
  57. xapproachesinfinity Group Title
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    8x^2/4x^2-1

    • 2 months ago
  58. cutelilgirl Group Title
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    yes but this seems a little beyond me :(

    • 2 months ago
  59. myininaya Group Title
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    have you talked about integration by parts?

    • 2 months ago
  60. myininaya Group Title
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    that is the only way i know how to do this problem

    • 2 months ago
  61. myininaya Group Title
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    so i assume you have had it mentioned in class

    • 2 months ago
  62. myininaya Group Title
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    is that a correct assumption

    • 2 months ago
  63. cutelilgirl Group Title
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    yes

    • 2 months ago
  64. xapproachesinfinity Group Title
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    can't do partial fraction here

    • 2 months ago
  65. myininaya Group Title
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    do you know how to evaluate \[\int \ln(x) dx ]

    • 2 months ago
  66. myininaya Group Title
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    \[\int\limits_{}^{} \ln(x) dx\]

    • 2 months ago
  67. myininaya Group Title
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    If you don't I can show you that may help you seeing this more simple integral evaluated first

    • 2 months ago
  68. xapproachesinfinity Group Title
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    we need another integration by part? i guess

    • 2 months ago
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