# Homework Help: Comprison test for Integrals

1. Apr 12, 2008

### SamJay

Hey guys. I'm a little new on the comparison test, if you could just check that i'm on the right track, it would be great.

1. The problem statement, all variables and given/known data
Using comparison test to decide whether or not $$\frac{\sqrt{x}}{\ln{x} + x^2}$$ converges.

2. Relevant equations
(Check 3.)

3. The attempt at a solution
Okay, so first off, I guess that this converges. So find a function that is larger than the one that is given and test if that converges, yes?

Would: $$\frac{\sqrt{x}}{\ln{x} + x^2} < \frac{\sqrt{x}}{x^2}$$ be a good choice?

So would I be right in saying that as $$x^{-3/2}$$ converges, that also the one given converges?

I'm not sure if i've gotten this backwards or not. Could someone please clear it up for me?

Thanks.

2. Apr 12, 2008

### HallsofIvy

First of all, it doesn't make any sense to ask "whether $$\frac{\sqrt{x}}{\ln{x} + x^2}$$ converges". Do you mean to determine whether or not an integral of that converges? If so then you still need to specify whether you are asking about the improper integral [tex]\int_a^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex] with a> 0 or [tex]\int_0^a \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex], again with a> 0.

3. Apr 12, 2008

### SamJay

... Whoops. So sorry, forgot to include the integral.

But yes, check if the Integral converges. Sorry about that.

[tex]\int_1^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex]

Not sure why, but my upper bound won't stay as infinity. (Today's the first time i've used LaTeX.)

Bounds are 1 to Infinity.

Oh, and whenever I said in the first post whether or not things converge, it's when taking the limit as x approaches infinity. Sorry about not making that clear in the first post.

Last edited: Apr 12, 2008
4. Apr 12, 2008

### Dick

Your choice of x^(-3/2) to compare with is a good one. And yes, it works. You might want to pound that point in by saying why the inequality is true.

5. Apr 12, 2008

### SamJay

Alright, thanks. :)

I'll add in an extra step saying that x^2 < ln(x) + x^2 and explaining it. Thanks.