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.

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.
