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

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What have you tried?
I will type in what I've done right now

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

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