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lambchamps

  • 3 years ago

\[\int\frac{ (lnx) ^{2} }{ x }dx\]

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  1. Algebraic!
    • 3 years ago
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    u= 2ln(x) du = 2/x

  2. lambchamps
    • 3 years ago
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    then

  3. Algebraic!
    • 3 years ago
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    integral of 1/2*u*du

  4. lambchamps
    • 3 years ago
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    and

  5. Algebraic!
    • 3 years ago
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    u srs?

  6. Algebraic!
    • 3 years ago
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    or should I say u^2/2 srs?

  7. lambchamps
    • 3 years ago
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    don't get it, because the answer is \[\frac{ 1 }{3 }lnx ^{3}+c\]

  8. lambchamps
    • 3 years ago
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    i just don't know how to solve it

  9. Algebraic!
    • 3 years ago
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    don't think you can write it like that...

  10. calculusfunctions
    • 3 years ago
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    Alright, do you know the substitution rule for integrals? Also do you know the derivatives of logarithm functions? If you do, then we're off to a great start!

  11. lambchamps
    • 3 years ago
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    yep

  12. calculusfunctions
    • 3 years ago
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    Were you replying to me or Algebraic?

  13. lambchamps
    • 3 years ago
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    you

  14. calculusfunctions
    • 3 years ago
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    OK, so then when you see\[\int\limits_{}^{}\frac{ \ln x ^{2} }{ x }dx\]What do you see as the first potential step?

  15. Callisto
    • 3 years ago
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    Sorry to interrupt, is the question (i) \(\int \frac{lnx^2}{x}dx\) or (ii) \(\int \frac{(lnx)^2}{x}dx\)? They are different...

  16. calculusfunctions
    • 3 years ago
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    Do you know your logarithm properties? for example\[\ln x ^{n}=n \ln x\]Do you know this property?

  17. Algebraic!
    • 3 years ago
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    yeah it's supposed to be (ln(x))^2

  18. calculusfunctions
    • 3 years ago
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    @Callisto, it's the former.

  19. Callisto
    • 3 years ago
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    For the first one, I don't think you can get (1/3)(lnx)^3 +C

  20. Algebraic!
    • 3 years ago
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    nope:)

  21. lambchamps
    • 3 years ago
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    turn it to \[\int\limits_{?}^{?}\frac{ 1 }{ x } \times \ln x ^{2} dx\]

  22. Algebraic!
    • 3 years ago
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    that clear it up @lambchamps ?

  23. lambchamps
    • 3 years ago
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    the second @Callisto

  24. Callisto
    • 3 years ago
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    You see... That's why...

  25. lambchamps
    • 3 years ago
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    sorry it is the second guys

  26. Algebraic!
    • 3 years ago
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    so it's u = ln(x) u^2 = (ln(x))^2 du = 1/x go to town.

  27. calculusfunctions
    • 3 years ago
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    @Callisto, you are right but @lambchamps, Is the question written correctly before we proceed further?

  28. lambchamps
    • 3 years ago
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    @calculusfunctions it's the second. based on what @Callisto have mentioned

  29. lambchamps
    • 3 years ago
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    @Algebraic! please continue

  30. calculusfunctions
    • 3 years ago
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    OK! so then we don't need the logarithm property I proposed earlier.

  31. Algebraic!
    • 3 years ago
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    that's it man, plug em in and integrate.

  32. calculusfunctions
    • 3 years ago
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    @lambchamps, do you now what the derivative of\[y =\ln x\]is?

  33. Algebraic!
    • 3 years ago
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    @calculusfunctions he or she is in calc. 2 so yeah probably. good question though.

  34. lambchamps
    • 3 years ago
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    dy = 1/x dx?

  35. calculusfunctions
    • 3 years ago
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    Great, so then if\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]then in order to apply the substitution rule what should u equal?

  36. lambchamps
    • 3 years ago
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    \[=\int\limits_{}^{}\frac{ 1 }{ x } \times \ln x ^{2} dx\]

  37. lambchamps
    • 3 years ago
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    i mean (ln x)^2

  38. calculusfunctions
    • 3 years ago
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    First of all you keep confusing the issue by writing\[\ln x ^{2}\]instead of\[(\ln x)^{2}\]They are not the same!\[(\ln x)^{2}\neq \ln x ^{2}\]Do you understand? So we're not going to get anywhere until this gaffe is resolved.

  39. calculusfunctions
    • 3 years ago
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    OH, OK! Sorry, I see you did fix it.

  40. lambchamps
    • 3 years ago
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    sorry it's \[(\ln x)^{2}\]

  41. calculusfunctions
    • 3 years ago
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    Do you notice that the derivative of ln x is in the integrand? So then what should u equal?

  42. lambchamps
    • 3 years ago
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    don't know please do tell

  43. lambchamps
    • 3 years ago
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    is it the \[\frac{ (\ln x)^{n+1} }{ n+1 } ?\]

  44. calculusfunctions
    • 3 years ago
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    Here are your options. Do you think u should equal a). ln x or b). 1/x c). (ln x)²

  45. calculusfunctions
    • 3 years ago
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    Which option do you think? a, b, or c? Just keep in mind that whichever option you choose, it's derivative must be in the integrand.

  46. lambchamps
    • 3 years ago
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    the derivative of ln x is 1/x so i choose b

  47. calculusfunctions
    • 3 years ago
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    NO! I said that the derivative of the chose option must be in the integrand. NOT the antiderivative of the option must be in the integrand.

  48. calculusfunctions
    • 3 years ago
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    So what should u equal?

  49. lambchamps
    • 3 years ago
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    a?

  50. calculusfunctions
    • 3 years ago
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    Excellent!

  51. lambchamps
    • 3 years ago
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    and then

  52. calculusfunctions
    • 3 years ago
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    So now\[u =\ln x\]so then\[du =?\]

  53. lambchamps
    • 3 years ago
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    1/x

  54. calculusfunctions
    • 3 years ago
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    \[du =\frac{ 1 }{ x }dx\]OK?

  55. lambchamps
    • 3 years ago
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    ok sorry

  56. calculusfunctions
    • 3 years ago
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    So now if you substitute u and du into your integral, what do you have?

  57. calculusfunctions
    • 3 years ago
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    @integralsabiti, how is giving the solution helping the student who is trying to learn? I spent all this time trying to teach @lambchamps so that she can then do other similar problems with confidence, and you just came in wasted her time and my effort. NOT COOL!

  58. lambchamps
    • 3 years ago
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    @calculusfunctions effort appreciated.. thanks to both of you

  59. integralsabiti
    • 3 years ago
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    sorry for interrupting .you may go on

  60. calculusfunctions
    • 3 years ago
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    @integralsabiti, thank you, no worries now that I know your intentions were genuine.

  61. calculusfunctions
    • 3 years ago
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    @lambchamps, are you still there? I'm still waiting for your response to my last question regarding your problem.

  62. lambchamps
    • 3 years ago
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    so I would need to find the dx then substitute the value to it?

  63. mikala1
    • 3 years ago
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    ok calculausfunction after your done here would you help me on my problem pleases

  64. calculusfunctions
    • 3 years ago
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    So what do you you now have after substituting\[u =\ln x\]and\[du =\frac{ 1 }{ x }dx\]into\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]

  65. calculusfunctions
    • 3 years ago
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    Of course but let's hurry and finish this one first because I have to log out soon.

  66. calculusfunctions
    • 3 years ago
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    So what does the integral look like after substitution?

  67. lambchamps
    • 3 years ago
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    dx would be xdu?

  68. calculusfunctions
    • 3 years ago
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    NO! If \[u =\ln x\]and\[du =\frac{ 1 }{ x }dx\]then\[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx =\int\limits_{}^{}u ^{2}du\]Do you see how?

  69. lambchamps
    • 3 years ago
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    so the 1/x in (ln x)^2/x would be cancelled?

  70. calculusfunctions
    • 3 years ago
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    @mikala, I'll help you right after I finish here.

  71. lambchamps
    • 3 years ago
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    because of the x.du

  72. lambchamps
    • 3 years ago
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    please tell me i'm right

  73. calculusfunctions
    • 3 years ago
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    No, the\[\frac{ 1 }{ x }dx\]is being replaced with du because\[du =\frac{ 1 }{ x }dx\]Please tell me you see that.

  74. calculusfunctions
    • 3 years ago
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    Cancelled is a poor choice of words.

  75. lambchamps
    • 3 years ago
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    ok, but where did the 1/x go? the one below (ln x)^2

  76. lambchamps
    • 3 years ago
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    oh i'm sorry didn't see at first

  77. calculusfunctions
    • 3 years ago
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    \[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]is exactly the same as\[\int\limits_{}^{}(\frac{ 1 }{ x }∙(\ln x)^{2})dx\]correct?

  78. lambchamps
    • 3 years ago
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    yes

  79. lambchamps
    • 3 years ago
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    i see the 1/x dx

  80. lambchamps
    • 3 years ago
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    after finding that, then is the time to integrate, right?

  81. calculusfunctions
    • 3 years ago
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    So I replaced the factor of\[\frac{ 1 }{ x }dx\]with du because\[du =\frac{ 1 }{ x }dx Yes, now you may integrate\[\int\limits_{}^{}u ^{2}du\]

  82. calculusfunctions
    • 3 years ago
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    Sorry, I don't know what happened there let me try again.

  83. lambchamps
    • 3 years ago
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    ok

  84. calculusfunctions
    • 3 years ago
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    Yes, now find the\[\int\limits_{}^{}u ^{2}du\]Can you do that please?

  85. lambchamps
    • 3 years ago
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    \[\frac{ 1 }{ 3 }u ^{3} + c\]

  86. calculusfunctions
    • 3 years ago
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    Right!

  87. calculusfunctions
    • 3 years ago
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    Now what's the final step?

  88. lambchamps
    • 3 years ago
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    then substitute

  89. lambchamps
    • 3 years ago
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    ln x right?

  90. calculusfunctions
    • 3 years ago
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    Yes so can you please write the final answer now?

  91. lambchamps
    • 3 years ago
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    \[\frac{ (\ln x)^{3} }{ 3 } + C\] ayt?

  92. calculusfunctions
    • 3 years ago
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    Perfect!!!

  93. lambchamps
    • 3 years ago
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    on more question is the c suppose to capitalized?

  94. lambchamps
    • 3 years ago
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    one*

  95. calculusfunctions
    • 3 years ago
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    That doesn't matter. C represents a constant. Whether you write c or C or k or K etc. is irrelevant. Just don't use x, y, or z.

  96. calculusfunctions
    • 3 years ago
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    The most commonly used ones are c and k.

  97. lambchamps
    • 3 years ago
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    thank you very much Sheldon Cooper you're the best

  98. calculusfunctions
    • 3 years ago
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    HAHAHA! @lambchamps, thank you very much! "The Big Bang Theory" is my favourite show!

  99. lambchamps
    • 3 years ago
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    i bet.. later dude

  100. calculusfunctions
    • 3 years ago
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    Later!

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