Determine if the integral converges or diverges using the Comparison Theorem (CT).
$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$
I saw solutions online arrive at the following inequality
$$\frac{x+1}{\sqrt{x^4 -x}} \ge \frac{1}{x}$$
But this inequality only holds for $x > 1$ and not for $x \ge 1$ since at $x=1$ the LHS of the inequality, $\frac{x+1}{\sqrt{x^4 -x}}$, is undefined.
I thought the inequality had to be true over the limits of integration, $x \ge 1$ or $[1, \infty]$ in this case, in order to use the CT.
Furthermore, the CT requires the functions being compared to be continuous over a closed interval, but at $x = 1$ function $\frac{x+1}{\sqrt{x^4 -x}}$ is infinitely discontinuous (vetical asymptote) which I'm assuming means we can't apply the CT over $[1, \infty]$.
But I have seen solutions online show that
$$\int_{1}^{\infty} \frac{1}{x}dx$$
diverges and thus $$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$ diverges by the CT. Are these solutions correct?
As a work around I was thinking the inequality is true for $x \ge 2$ so the integral could be "split up" into the following
$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx = \int_{1}^{2} \frac{x+1}{\sqrt{x^4 -x}}dx + \int_{2}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$
The CT could then be applied to the second integral. The first integral, however, still has an infinite discontinuity at $x = 1$ and thus is an "improper" integral. This would require evaluating the integral directly which defeats the purpose of using the CT in the first place to determine convergence or divergence.
My questions are:
Does it matter if the inequality is not true over the interval or limits of integration $[a, \infty)$ or $x \ge a$ in order to apply the Comparison Theorem?
Were the solutions online correct?