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jotopia34 Group Title

What is the integral of ln(sqrt t)/t, stuck again!

  • one year ago
  • one year ago

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  1. tkhunny Group Title
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    What have you tried?

    • one year ago
  2. jotopia34 Group Title
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    I will type in what I've done right now

    • one year ago
  3. tkhunny Group Title
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    Sweet!

    • one year ago
  4. amoodarya Group Title
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    • one year ago
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  5. tkhunny Group Title
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    ?? Is that the same problem?

    • one year ago
  6. jotopia34 Group Title
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    \[u=\sqrt{t}\] du= 1/2 t^-1/2 dt du=1/2*1/sqr(t) dt But I know its wrong because I still can't cancel anything

    • one year ago
  7. amoodarya Group Title
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    i think it was integral ln sqrt x dx isnt it ? or it is int ln(sqrtx)/x dx ?

    • one year ago
  8. tkhunny Group Title
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    Why would you use a square root substitution when you can use a logarithm property?

    • one year ago
  9. slaaibak Group Title
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    yeah.. ln(sqrt(t)) = 1/2 ln t

    • one year ago
  10. jotopia34 Group Title
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    what is that property again please? in general format?

    • one year ago
  11. slaaibak Group Title
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    \[\ln(x^a) = a \times \ln(x)\]

    • one year ago
  12. jotopia34 Group Title
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    ohhhh, now I see, the square root is 1/2 as exponent. Thus, that is my (a). Thank u!!

    • one year ago
  13. slaaibak Group Title
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    Cool :) do you know how to go from here?

    • one year ago
  14. jotopia34 Group Title
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    yes, I believe so

    • one year ago
  15. slaaibak Group Title
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    shot

    • one year ago
  16. jotopia34 Group Title
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    is the answer 1/4 * (lnx)^2+C? I think its wrong. I saw a post from another guy and I don't get what he wrote. Slaaibak, you make excellent sense. Can you post the solution steps? Id be eternally grateful...

    • one year ago
  17. amoodarya Group Title
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    • one year ago
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  18. jotopia34 Group Title
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    Amoodarya, your last posting is the final answer I got, so I think I got it right....

    • one year ago
  19. slaaibak Group Title
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    Sure. \[\int\limits {1 \over 2} {\ln t \over t} dt\] Let's set k = ln t dk = 1/t dt so the integral becomes; \[{1 \over 2} \int\limits k dk = {1 \over 4 }k^2 + C = {1 \over 4}(\ln t)^2 + C\]

    • one year ago
  20. jotopia34 Group Title
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    yes!! You are all geniuses!

    • one year ago
  21. slaaibak Group Title
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    haha, if only :( hope it helped. thank you for the fun, makes insomnia a bit better.

    • one year ago
  22. jotopia34 Group Title
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    insomia is the curse of genius

    • one year ago
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