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jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0I will type in what I've done right now

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0?? Is that the same problem?

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0\[u=\sqrt{t}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0i think it was integral ln sqrt x dx isnt it ? or it is int ln(sqrtx)/x dx ?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Why would you use a square root substitution when you can use a logarithm property?

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.1yeah.. ln(sqrt(t)) = 1/2 ln t

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0what is that property again please? in general format?

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.1\[\ln(x^a) = a \times \ln(x)\]

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0ohhhh, now I see, the square root is 1/2 as exponent. Thus, that is my (a). Thank u!!

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.1Cool :) do you know how to go from here?

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0is 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...

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0Amoodarya, your last posting is the final answer I got, so I think I got it right....

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.1Sure. \[\int\limits {1 \over 2} {\ln t \over t} dt\] 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
 one year ago
Best ResponseYou've already chosen the best response.0yes!! You are all geniuses!

slaaibak
 one year ago
Best ResponseYou've already chosen the best response.1haha, if only :( hope it helped. thank you for the fun, makes insomnia a bit better.

jotopia34
 one year ago
Best ResponseYou've already chosen the best response.0insomia is the curse of genius
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