Answer
Diverges
Work Step by Step
$\displaystyle \int_0^1\frac{\sec^2x}{x\sqrt{x}}\ dx = \int_0^1\frac{dx}{x\sqrt{x} \cos^2x}$
Find a lesser-valued function:
$\displaystyle \frac{1}{x\sqrt{x} \cos^2x} > \frac{1}{2x\sqrt{x} } = \frac{x^{-\frac{3}{2}}}{2}$
(Recall: $\displaystyle 0 \le \cos x \le 1$, so substitute with 2 as this will make the value lower since it is in denominator)
$\displaystyle \frac{1}{2}\int_0^1x^{-\frac{3}{2}}\ dx = \lim\limits_{t \to 0} \frac{1}{2} \int_t^1x^{-\frac{3}{2}}\ dx$
$\displaystyle = \frac{1}{2}\lim\limits_{t \to 0} \left[ -2x^{-\frac{1}{2}} \right]_t^1$
$\displaystyle = -\lim\limits_{t \to 0} \left[ 1-\frac{1}{\sqrt{t}}) \right]$
$\displaystyle = - \left[ 1-\infty \right]$
Divergent by the comparison theorem, since the lesser-valued function diverges.
