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wafaa
Group Title
evaluate s x( (8+2xx^2) )^1/2 dx
p/s: s is for integrate
 one year ago
 one year ago
wafaa Group Title
evaluate s x( (8+2xx^2) )^1/2 dx p/s: s is for integrate
 one year ago
 one year ago

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Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
I assume its an integral \[\int\limits x\sqrt{8+2xx^{2}}dx \] Well, complete the square first; then looks like a sub
 one year ago

wafaa Group TitleBest ResponseYou've already chosen the best response.0
yes. I've done the completing the square but not sure whats the next step. by part or substitution?
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
looks like a substitution would be easier
 one year ago

wafaa Group TitleBest ResponseYou've already chosen the best response.0
ok. i"ll try again. tq :)
 one year ago

wafaa Group TitleBest ResponseYou've already chosen the best response.0
still dont get it. help me plez :(
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits x\sqrt{x^{2}+2x+8} dx => \int\limits x\sqrt{(x^{2}2x8)} dx => \int\limits x\sqrt{((x^{2}1)^{2}(1)^{2}+8))} dx\] \[=> \int\limits x\sqrt{(x^{2}1)9} dx => \int\limits x\sqrt{9(x1)^{2}} dx\] let u = x1 > x = u+1 \[=> \int\limits\left(u+1\right)\sqrt{9u^{2}}du \]
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
then it's trig sub
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[ x= sin\theta\]
 one year ago

wafaa Group TitleBest ResponseYou've already chosen the best response.0
ok mimi.i'll try. thanks again.
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
np and for this u = 3sin\theta
 one year ago
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