Improper integral comparison test

[hitemup]

[tex]\int_{0}^{\infty} \frac{x^2 dx}{x^5+1}[/tex]
The question asks whether this function diverges or converges.
I have tried to do some comparisons with x^2/(x^6+1), and x^2/(x^3+1) but it didn't end up with something good.

Then I decided to compare it with [tex] \frac{x^2}{x^4+1}[/tex]
Since this function converges and is greater than the given function on [tex](1,\infty )[/tex] it proves that the given function converges too. But it almost takes one page to integrate this function so I thought there must be an easier way to handle this. What other function can I think of rather than this?

 

[hitemup]

I think I've found it.

[tex]\frac{x^2}{x^5}[/tex] is greater than the given function. Can I do comparison test between these two on (1, infinity)? Then I add the rectangle formed by x = 0 y =0 and x= 1 y=1/2 which comes from the given function. Sum of these two areas must be finite and greater than x^2/(x^5+1), so it proves asked expression is convergent, doesn't it?

 

[Mark44]

I think I've found it.

[tex]\frac{x^2}{x^5}[/tex] is greater than the given function.

This is the most obvious one to use for comparison. It should have been your first choice, but it takes some practice to be able to notice things like this right away. Note that x²/x⁵ is the same as 1/x³.

Can I do comparison test between these two on (1, infinity)?

I don't see anything wrong with that. Your original integrand is defined on the interval [0, 1], so it's easy enough to evaluate the integral using those limits. Then you can both integrals on [1, ∞).

Then I add the rectangle formed by x = 0 y =0 and x= 1 y=1/2 which comes from the given function. Sum of these two areas must be finite and greater than x^2/(x^5+1), so it proves asked expression is convergent, doesn't it?