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does the integral of 1/x+ srqtx from 1 to infinity converge?

Mathematics
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i doubt it
the degree of the denominator has to be larger than the degree of the numerator by more than 1
is it \[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\] OR \[\int_1^\infty{dx\over x+\sqrt{x}} \]

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Other answers:

the second one
@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!!
so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously
oh no!!
oh yes! :D
just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough
simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge
neither does the one above
aah.. 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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