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You're taking the integral of such function for all $x\ge1$ and within this range, the function is continuous as said above which is important to bound the integrand the way we want. So for example $\dfrac{1}{{{x}^{6}}+2}<\dfrac{1}{{{x}^{6}}}$ holds always for $x\ge1$ and besides $3x^2-2<3x^2+2$ holds always, then actually for all $x\ge1$ you have \(\displaystyle \displaystyle\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}<\frac{3{{x}^{3}}+2}{{{x}^{6}}}= \frac{3}{x^3}+\frac{2}{{{x}^{6}}},\) implying $\displaystyle\int_{1}^{\infty }{\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}\,dx}<\infty $Hey, not too sure about what function i would compare this integral from 1 to infinity of (3x^3 -2)/(x^6 +2) dx. I also have to show that it converges.