Question

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

Answer 1

i doubt it

Answer 2

the degree of the denominator has to be larger than the degree of the numerator by more than 1

Answer 3

is it
\[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\]
OR
\[\int_1^\infty{dx\over x+\sqrt{x}}
\]

Answer 4

the second one

Answer 5

@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!!

Answer 6

so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously

Answer 7

oh no!!

Answer 8

oh yes! :D

Answer 9

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

Answer 10

simple example
\[\int_1^{\infty}\frac{dx}{x}\] does not converge

Answer 11

neither does the one above

Answer 12

aah.. jumping to conclusions too soon.
\[
u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\
\int_1^\infty\frac{2u}{u^2+u}du\\
=2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty}
\]