Answer
The integral diverges.
Work Step by Step
$\displaystyle \int^{\infty}_{1} \frac{x+1}{\sqrt{x^4-x}}\ dx$
We have:
$\displaystyle \frac{x+1}{\sqrt{x^4-x}} > \frac{x}{\sqrt{x^4-x}} > \frac{x}{\sqrt{ x^4}} = \frac{x}{x^2} = \frac{1}{x}$
From the textbook, we know that $\displaystyle \int^{\infty}_{1} \frac{1}{x^p}\ dx$ diverges when $\displaystyle p \le 1$.
Since the lesser-valued function $\displaystyle \int^{\infty}_{1} \frac{1}{x}\ dx$ diverges, then by the comparison theorem, $\displaystyle \int^{\infty}_{1} \frac{x+1}{\sqrt{x^4-x}}\ dx$
diverges.
