Does x/(√(1+x⁶)) converge using comparison test?

[mpgcbball]

Hi, I'm having a bit of trouble showing that the integral from 1 to infinity of x/((1+x^6))^1/2 converges or diverges by the comparative property.

I'm not sure if I'm setting it up right, but so far I have that 1/rad(1+x^6) is less than or equal to x/rad(1+x^6) which is less than 1/rad(x^6).

I don't know if this is right, or where to go from here if it is right.

Thanks for your help!

 

[Dick]

Do you think it converges or diverges? What is your gut feeling. If x is large what does it look like (ie ignoring the 1)?

 

[Gib Z]

$$\sqrt{x^6+1}>\sqrt{x^6}$$. (Trivial)$$\frac{x}{\sqrt{x^6+1}}<\frac{x}{\sqrt{x^6}}$$
For every value of x within our bounds of integration, that is true.
$$\int^{\infty}_1 \frac{x}{\sqrt{x^6+1}} dx< \int^{\infty}_1 \frac{1}{x^2} dx$$. Since we know, and can show, the 2nd part converges, the 1st part does as well.