## anonymous 3 years ago does the integral of 1/x+ srqtx from 1 to infinity converge?

1. anonymous

i doubt it

2. anonymous

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

3. anonymous

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

4. anonymous

the second one

5. anonymous

@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!!

6. anonymous

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

7. anonymous

oh no!!

8. anonymous

oh yes! :D

9. anonymous

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

10. anonymous

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

11. anonymous

neither does the one above

12. anonymous

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