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kugler97 Group TitleBest ResponseYou've already chosen the best response.0
what is the integer?
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{} \frac{ (2t+1)^{2} }{ \sqrt{t} } dt\]
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
does \[\frac{ t ^{2} }{ \sqrt{t} } \] cancel out????
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
u need to expand of (2t + 1)^2 = 4t^2 + 4t + 1 and convert sqrt(t) = t^(1/2) now, can u simplify (4t^2 + 4t + 1)/t^(1/2) ?
 one year ago

kugler97 Group TitleBest ResponseYou've already chosen the best response.0
yea i think it does lol
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
well I'll show you how far I got \[4 \int\limits_{}^{} \frac{ t ^{2} }{ t ^{1/2} } +4 \int\limits_{}^{} \frac{ t }{ t ^{1/2} } + \int\limits_{}^{} \frac{ 1 }{ t ^{1/2} }\]
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
so far is good, then use the properties of exponent to simplify of them : dw:1358987039172:dw
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
now, what would be ur integral ?
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
Sorry openstudy crapped out last night lol thanks
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
nopes.. :) btw, do u got it ?
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
Mhmm :) I got \[\frac{ 8t ^{^{5/2}} }{ 5 }+\frac{ 8t ^{3/2} }{ 3 }+ 2t ^{1/2}+c\] Look about right?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
yeah, i have cheked ur answer you are right :)
 one year ago

KatClaire Group TitleBest ResponseYou've already chosen the best response.0
Hehe you are the best thanks!!!!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.3
you're welcome
 one year ago
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