integral of 1/(sqrt(x)+x^2) from 0 to infinity

[cheffy]

Does the
integral of 1/(sqrt(x)+x^2) from 0 to infinity
converge or diverge?

I split the integral into two, from 0 to 1 and from 1 to infinity. I know that 0 to 1 converges, but I can't figure out how to do from 1 to infinity. Any suggestions?

Thanks!

 

[pka]

You do know that if you could post using TeX, we could read your posting:?
Go to the top of this page: "FORUM HELP".
Find there a 'help' session on TeXt.

 

[soroban]

Hello, cheffy!

I think I read the problem correctly . . .

Does \(\displaystyle \L\:\int^{\;\;\;\infty}_0\frac{dx}{\sqrt{x}\,+\,x^2}\) converge or diverge?

I split the integral into two, from 0 to 1 and from 1 to infinity. . Why?
I know that 0 to 1 converges. . How?

Did you try to integrate it?

Let \(\displaystyle u\,=\,\sqrt{x}\;\;\Rightarrow\;\;x\,=\,u^2\;\;\Rightarrow\;\;dx\,=\,2u\,du\)

Substitute: \(\displaystyle \L\:\int\frac{2u\,du}{u\,+\,u^4}\;=\;2\int\frac{du}{u^3\,+\,1}\;=\;2\int\frac{du}{(u\,+\,1)(u^2\,-\,u\,+\,1)}\)

This can be integrated, but it's not pretty . . .


Partial fractions: \(\displaystyle \L\:\frac{2}{3}\int\left[\frac{1}{u\,+\,1}\,-\,\frac{1}{u^2\,-\,u\,+\,1}\,+\,\frac{2}{u^2\,-\,u\,+\,1}\right]\,du\)

. . . . . and so on.