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satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the degree of the denominator has to be larger than the degree of the numerator by more than 1

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0is it \[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\] OR \[\int_1^\infty{dx\over x+\sqrt{x}} \]

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 aha... @peggiepenguin next time please use the "equation editor on this website" or "TEX" to enter your equations.. so we know what you are talking about!!

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1neither does the one above

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0aah.. jumping to conclusions too soon. \[ u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\ \int_1^\infty\frac{2u}{u^2+u}du\\ =2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty} \]
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