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Indefinite integratrion

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    integrate lnsqrt(t)/t , dt


    2. Relevant equations



    3. The attempt at a solution

    This is the most i can come up with.

    u = lnsqrt(t), so du = 1/(2sqrt(t))dt , so dt = 2du/sqrt(t)

    Im stuck @ this point
     
  2. jcsd
  3. Sep 21, 2010 #2
    Hint: The logarithm function has an interesting property.

    [tex] ln\left(a^{b} \right) = bln\left(a \right)[/tex]
     
  4. Sep 21, 2010 #3

    Dick

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    du isn't 1/(2*sqrt(t))*dt. You are forgetting to use the chain rule. You can also get the correct differential by applying the rules of logs to ln(sqrt(t))=ln(t^(1/2)).
     
  5. Sep 21, 2010 #4
    so i would integrate (lnt^(1/2))/t ?
     
  6. Sep 21, 2010 #5

    Dick

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    Sure. Isn't ln(sqrt(t))/t=ln(t^(1/2))/t?
     
  7. Sep 21, 2010 #6
    Ok, now i'm not sure what to take as u.
     
  8. Sep 21, 2010 #7
    Yes.
    But follow Dicks' suggestion also. It would be helpful to you. Your original method was correct ,however, as already pointed out you weren't differentiating correctly.

     
  9. Sep 21, 2010 #8

    Dick

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    ??? Take u the same as you originally chose. u=ln(sqrt(t))=ln(t^(1/2)). They are the same thing. You just have to compute du correctly. That's where your first attempt went wrong. The du was wrong.
     
  10. Sep 21, 2010 #9
    Yea i think i integrated rather then differentiated to get du. so u = 1/2lnt , du = 1/2(1/t)dt?
     
  11. Sep 21, 2010 #10
    actually du = 1/t
     
  12. Sep 21, 2010 #11
    that would make give me 1/2lnt (1/t), which is udu +C. so it breaks down to u^2/2 + C?
     
  13. Sep 21, 2010 #12

    MysticDude

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    Last edited by a moderator: Apr 25, 2017
  14. Sep 21, 2010 #13

    Dick

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    Sure. You could have also done it with ln(sqrt(t)). (ln(sqrt(t))'=(1/sqrt(t))*(dsqrt(t)/dt)=(1/sqrt(t))*(1/(2*sqrt(t)))=1/(2t). That's the chain rule. BTW if you aren't going to use TeX you should probably use a few more parentheses. 1/2lnt could mean (1/2)ln(t) or 1/(2ln(t)) etc.
     
  15. Sep 21, 2010 #14

    MysticDude

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    Okay I'll think about that next time.
     
  16. Sep 21, 2010 #15
    Last edited by a moderator: Apr 25, 2017
  17. Sep 21, 2010 #16

    MysticDude

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    No all it is is just normal differentiation. See all you have to do if factor out the constant( 1/2 in this case). So it would look like [tex](\frac{1}{2}) * (\frac{d}{dt})(ln(t)) = \frac{1}{2t}[/tex]
     
  18. Sep 21, 2010 #17
    but i thought the 1/2 would be eliminated because it is a constant
     
  19. Sep 21, 2010 #18

    MysticDude

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    If you can factor out a constant, then do so. Plus if you look at the link that I gave you. There should be a "Show Steps" link. It says to factor out the constant.

    AND even if you do it by using the product rule (I'm doing this so you can see that the 1/2 stays)

    [tex](\frac{1}{2}[(\frac{d}{dt})(ln(t))] + (\frac{d}{dt})(\frac{1}{2})(ln(t))[/tex]
    We have [tex]\frac{1}{2}*\frac{1}{t} + 0(this-is-the-derivative-of \frac{1}{2}) * ln(t)[/tex]
    so this shows that the derivative of [tex]\frac{1}{2} * ln(t) = \frac{1}{2t}[/tex]


    I hope you understand!
     
  20. Sep 21, 2010 #19

    Thank you
     
  21. Sep 21, 2010 #20

    Dick

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    Ok, so after all that fuss, you can do the integral, right?
     
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