I'm trying to use the comparison theorem to show that $\displaystyle \int_1^{\infty}\frac{x+1}{\sqrt{x^4-x}}dx$ is divergent.

By graphing the integrand, it appears that $\displaystyle 0\leq\frac{1}{x}\leq\frac{x+1}{\sqrt{x^4-x}}$ throughout the interval $\displaystyle (1,\infty]$. Since $\displaystyle \int_1^{\infty}\frac{dx}{x}=\lim_{t->\infty}\left(\ln(t)-\ln(1)\right)$ diverges, does this prove that $\displaystyle \int_1^{\infty}\frac{x+1}{\sqrt{x^4-x}}dx$ diverges? Is it that easy?