jotopia34
  • jotopia34
What is the integral of ln(sqrt t)/t, stuck again!
Mathematics
jamiebookeater
  • jamiebookeater
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tkhunny
  • tkhunny
What have you tried?
jotopia34
  • jotopia34
I will type in what I've done right now
tkhunny
  • tkhunny
Sweet!

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amoodarya
  • amoodarya
1 Attachment
tkhunny
  • tkhunny
?? Is that the same problem?
jotopia34
  • jotopia34
\[u=\sqrt{t}\] 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
amoodarya
  • amoodarya
i think it was integral ln sqrt x dx isnt it ? or it is int ln(sqrtx)/x dx ?
tkhunny
  • tkhunny
Why would you use a square root substitution when you can use a logarithm property?
slaaibak
  • slaaibak
yeah.. ln(sqrt(t)) = 1/2 ln t
jotopia34
  • jotopia34
what is that property again please? in general format?
slaaibak
  • slaaibak
\[\ln(x^a) = a \times \ln(x)\]
jotopia34
  • jotopia34
ohhhh, now I see, the square root is 1/2 as exponent. Thus, that is my (a). Thank u!!
slaaibak
  • slaaibak
Cool :) do you know how to go from here?
jotopia34
  • jotopia34
yes, I believe so
slaaibak
  • slaaibak
shot
jotopia34
  • jotopia34
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...
amoodarya
  • amoodarya
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jotopia34
  • jotopia34
Amoodarya, your last posting is the final answer I got, so I think I got it right....
slaaibak
  • slaaibak
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; \[{1 \over 2} \int\limits k dk = {1 \over 4 }k^2 + C = {1 \over 4}(\ln t)^2 + C\]
jotopia34
  • jotopia34
yes!! You are all geniuses!
slaaibak
  • slaaibak
haha, if only :( hope it helped. thank you for the fun, makes insomnia a bit better.
jotopia34
  • jotopia34
insomia is the curse of genius

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