Finding if improper integral is converging or diverging

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.

• It is $O(x^{-3})$ for large $x$, and non-negative and bounded for $x\ge 1$, so converges – Henry Sep 19 '17 at 9:32

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.
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
• The integrand itself is equivalent to $\dfrac1{x3}$ near $\infty$. – Bernard Sep 19 '17 at 9:59
• 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$\dfrac{x^3}{x^6}$. – Bernard Sep 19 '17 at 10:11