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Integral problem with natural log ($5 to whoever gets it)

  1. Sep 9, 2004 #1
    Here's the problem:

    find the integral of


    In other words, dx over 2x times the square root of lnx.

    It has to be evaluated from 16 and 2. I don't know how to say it, but the integral sign with 16 at the top, and 2 at the bottom.

    If you can find the answerm and show me the work, preferably two different ways, I will paypal you $5.

    The sooner the better.

  2. jcsd
  3. Sep 9, 2004 #2


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    Welcome to physicsforums, formula107.

    First, a bit about our rules:

    1) Homework help posts should go into the appropriate homework help forums.
    2) You should show us the work you've attempted so far when asking for help.
    3) You should never offer money in return for our help.

    I assume you mean this integral?

    [tex]\int_2^{16} \frac{dx}{2 x (\ln x)^{1/2}}[/tex]

    - Warren
  4. Sep 9, 2004 #3
    Thanks for the advice. Yep, that was the integral I was trying to describe.

    I have attempted it, but I still get confused on what to do with dx on the top.

    One more question:
    Can I use the same text you are by using the [tex] code?

  5. Sep 9, 2004 #4


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    Can you please show us the paths you have already attempted to follow to find this integral? We do not generally just give away answers.
    Most certainly. You can click on any mathematical image to see the code that generated it. You can then copy and paste the code into your own posts, editing the code if you want. You can also quote a post to use or modify the code for the equations in that post.

    - Warren
  6. Sep 9, 2004 #5
    I tried bringing the bottom part to the top, so it makes

    [tex]\int_2^{16} \dx (2 x (\ln x)^{1/2})^{-1}[/tex]

    With dx on the end, but I can't get that to show.

    Then from there I just get completely lost. Is that a good first step?
    Last edited: Sep 9, 2004
  7. Sep 9, 2004 #6
    come on, this is darn easy, donate that $5 to charity.

    (Hint: Let [tex]y = \ln x[/tex])
  8. Sep 9, 2004 #7
    Yep, I finally got it. I end up with u^1/2, and with the numbers plugged in, I got (ln(2))^(1/2).

    Seems right to me!

    Just one more question: Is there any way to solve this problem without solving for dx before integrating. I came up with dx=xdu, since du=1/x, but we haven't done this in class yet. Is there a way to solve this using only u and du?

  9. Sep 10, 2004 #8
    If you do a u-sub, and let u be (lnx)^1/2 it turns into just the intergral of U*DU.
  10. Sep 11, 2004 #9
    So who ended up with the 5 dollars?
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