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tkhunny Group TitleBest ResponseYou've already chosen the best response.0
What have you tried?
 one year ago

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

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

jotopia34 Group TitleBest 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
 one year ago

amoodarya Group TitleBest ResponseYou've already chosen the best response.0
i think it was integral ln sqrt x dx isnt it ? or it is int ln(sqrtx)/x dx ?
 one year ago

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

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

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

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

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

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

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
yes, I believe so
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
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

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

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
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

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
yes!! You are all geniuses!
 one year ago

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

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