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Derivative of ln squared

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to understand how they got du in this problem. It is an integration by parts question. Anyway, the question asks me to evaluation this integral by using integration by parts two times
    Screen shot 2010-04-18 at 1.56.22 PM.png

    2. Relevant equations
    Screen shot 2010-04-18 at 2.06.04 PM.png
    This is how we would solve the problem...

    3. The attempt at a solution
    This is the first step.
    Screen shot 2010-04-18 at 1.55.51 PM.png

    The solution to the entire problem isn't what I need. I'll be able to get that after I figure out how they got du. I just need to see how you get du.

    What it looks like they did was just go 2*(ln x^21)*(1/x)*(21x^2)
    This isn't equal to the answer, though. Any ideas? No need to rush on this. The assignment is done, I just got monumentally frustrated with this one.

    Thanks in advance to anyone that can help.
  2. jcsd
  3. Apr 18, 2010 #2
    you suppose to get

    [tex]\frac{du}{dx}[/tex] = 42(ln x21)x20
  4. Apr 18, 2010 #3
    Is the substitution really necessary, I mean, is it part of what you're being taught right now and must use it?

    Because I find the following way to be more straightforward. Use integration by parts two times via this formula: [itex]\int f'(x)g(x)dx = f(x)g(x) - \int f(x)g'(x)dx[/itex]

    by doing the following: [itex]\int ln^2(x^{21})dx=\int (x)'ln^2(x^{21})dx=xln^2(x^{21}) - \int x(ln^2(x^{21}))'dx...[/itex]
  5. Apr 18, 2010 #4
    but that what we learn at school, the substitution is merely to make student more understand i guess. But ones you get it, you don't need those substitution.
  6. Apr 18, 2010 #5

    D H

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    Wrong. Look at the attachment to the original post. The correct answer is right there.

    So, to help with the original post.
    Did you do that?

    They gave you part of the first step. In particular, the attachment tells you exactly what to use how to express that integral in the form of [itex]\int u dv[/itex] needed for application of the integration by parts. Then it tells you what [itex]du[/itex] and [itex]v[/itex] are given that assignment. It does not perform the actual integration by parts, that is up to you. Then, as suggested, you should integrate by parts again, obviously with a different choice for u and dv this second step.
  7. Apr 18, 2010 #6
    owho, sorry, i missed to differentiate something, hoho, sorry sorry
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