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The question is to state whether the following integral diverges or converges: $$\int^{\infty}_{1} \frac{x^3-1}{\sqrt{x^{12}+x-1}}\, \mathrm dx$$

How do I go about finding this? I can't compute the integral, however I know that the denominator is always positive within the bounds, but I don't know what else to do.

All help is appreciated, thanks in advance.

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    $\begingroup$ It is $O(x^{-3})$ for large $x$, and non-negative and bounded for $x\ge 1$, so converges $\endgroup$ – Henry Sep 19 '17 at 9:32
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Since$$\lim_{x\to+\infty}\frac{\frac{x^3-1}{\sqrt{x^{12}+x-1}}}{\frac1{x^3}}=\lim_{x\to+\infty}\frac{x^6-x^3}{\sqrt{x^{12}+x-1}}=\lim_{x\to+\infty}\frac{1-\frac1{x^3}}{\sqrt{1+\frac1{x^{11}}-\frac1{x^{12}}}}=1$$and since the integral $\displaystyle\int_1^{+\infty}\frac1{x^3}\,\mathrm dx$ converges, your integral converges.

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What about looking at the integrand for sufficiently large $x$, it is always positive and furthermore $$\frac{x^3-1} {\sqrt{x^{12} + x - 1}} < \frac{x^3-1} {\sqrt{x^{12} }} =\frac{x^3-1} {x^{6} } \sim \frac{1}{x^3} $$ so the rate is fast enough for convergence

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  • $\begingroup$ The integrand itself is equivalent to $\dfrac1{x3}$ near $\infty$. $\endgroup$ – Bernard Sep 19 '17 at 9:59
  • $\begingroup$ @Bernard, sorry I fail to get the your point, could you please clarify? My little calc is indeed aimed at showing your claim. Of course one could skip the inequality and get the conclusion in one step, is that what you intend? $\endgroup$ – An aedonist Sep 19 '17 at 10:03
  • $\begingroup$ I just mean you can say directly that $\sqrt{x^{12}+x-1}\sim_\infty\sqrt{x^{12}}=x^6$, so the fraction is equivalent to$\dfrac{x^3}{x^6}$. $\endgroup$ – Bernard Sep 19 '17 at 10:11
  • $\begingroup$ Indeed, it was redundant, just wanted to make it clearer, point taken $\endgroup$ – An aedonist Sep 19 '17 at 10:14
  • $\begingroup$ Perhaps my point of view is biased: I tend to think there must be some intellectual effort from the question asker if we want the answer to be a little more than just the answer. Of course, if the O.P. asks for some clarification, I add details. $\endgroup$ – Bernard Sep 19 '17 at 10:18

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