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cutelilgirl

  • 4 months ago

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

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  1. satellite73
    • 4 months ago
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    start by factoring

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

  3. Elsa213
    • 4 months ago
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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

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

  5. Elsa213
    • 4 months ago
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    ik ik

  6. Elsa213
    • 4 months ago
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    it was just an example

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

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

  9. satellite73
    • 4 months ago
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    right

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

  11. xapproachesinfinity
    • 4 months ago
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    you forgot log lol

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

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

  14. cutelilgirl
    • 4 months ago
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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 ?

  15. xapproachesinfinity
    • 4 months ago
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    yes continue this way

  16. cutelilgirl
    • 4 months ago
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    ok thanks @xapproachesinfinity and @satellite73

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

  18. satellite73
    • 4 months ago
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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

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

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

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

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

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

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

  25. xapproachesinfinity
    • 4 months ago
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    yes i doesn't :)

  26. xapproachesinfinity
    • 4 months ago
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    it*

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

  28. satellite73
    • 4 months ago
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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

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

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

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

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

  33. xapproachesinfinity
    • 4 months ago
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    but you excluded the negative values?

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

  35. xapproachesinfinity
    • 4 months ago
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    wolfram is crazy hehe

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

  37. satellite73
    • 4 months ago
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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

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

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

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

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

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

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

  44. myininaya
    • 4 months ago
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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) \[=x \ln(f(x))-\int\limits_{}^{}x \frac{f'(x)}{f(x)} dx\]

  45. xapproachesinfinity
    • 4 months ago
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    so far it looks nice to me

  46. myininaya
    • 4 months ago
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    where did you get that @xapproachesinfinity ?

  47. xapproachesinfinity
    • 4 months ago
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    |dw:1408559618618:dw|

  48. xapproachesinfinity
    • 4 months ago
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    from by part integration

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

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

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

  52. xapproachesinfinity
    • 4 months ago
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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

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

  54. xapproachesinfinity
    • 4 months ago
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    yup you are right forgot that xD

  55. xapproachesinfinity
    • 4 months ago
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    it should be what you wrote above

  56. myininaya
    • 4 months ago
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    @cutelilgirl are you still there?

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

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

  59. myininaya
    • 4 months ago
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    have you talked about integration by parts?

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

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

  62. myininaya
    • 4 months ago
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    is that a correct assumption

  63. cutelilgirl
    • 4 months ago
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    yes

  64. xapproachesinfinity
    • 4 months ago
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    can't do partial fraction here

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

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

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

  68. xapproachesinfinity
    • 4 months ago
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    we need another integration by part? i guess

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