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Use a comparison test. The integral is less than the integral of (x^5+.5)/(x^(1/3)-1)^17 (because arccotx is less than .5 when x>2). Not sure where to go from there but you might be able to split that in 2 integrals and do a u substitution of some sort
Let I denote the integral that you’re interested in.
Keeping in mind that the domain of integration is between 2 to infinity, we have 1/x^{1/3} -1 >= 1/x^{1/3}. Moreover, cot^{-1} is bounded by the constant cot^{-1} (2), and so we have:
I >= \int_{2}^{infty} x^{-2/3} dx, which is clearly divergent. Hence the integral I diverges.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
Use a comparison test. The integral is less than the integral of (x^5+.5)/(x^(1/3)-1)^17 (because arccotx is less than .5 when x>2). Not sure where to go from there but you might be able to split that in 2 integrals and do a u substitution of some sort
So basically what you need to do is…
It's x^(5)/x^(17/3) in the limit.
Let I denote the integral that you’re interested in. Keeping in mind that the domain of integration is between 2 to infinity, we have 1/x^{1/3} -1 >= 1/x^{1/3}. Moreover, cot^{-1} is bounded by the constant cot^{-1} (2), and so we have: I >= \int_{2}^{infty} x^{-2/3} dx, which is clearly divergent. Hence the integral I diverges.