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

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

Mimi_x3
 2 years ago
Best ResponseYou've already chosen the best response.1looks like a substitution would be easier

wafaa
 2 years ago
Best ResponseYou've already chosen the best response.0ok. i"ll try again. tq :)

wafaa
 2 years ago
Best ResponseYou've already chosen the best response.0still dont get it. help me plez :(

Mimi_x3
 2 years ago
Best 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 \[=> \int\limits\left(u+1\right)\sqrt{9u^{2}}du \]

wafaa
 2 years ago
Best ResponseYou've already chosen the best response.0ok mimi.i'll try. thanks again.

Mimi_x3
 2 years ago
Best ResponseYou've already chosen the best response.1np and for this u = 3sin\theta
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