# Comparison Theorem

Use the Comparison Theorem to determine whether the integral is convergent/divergent: $\int_{0}^{1} \frac{sec^{2}x}{x\sqrt{x}} dx$
The secant function is bounded on the interval, so the convergence doesn't depend on that function. Close to zero, the numerator looks like 1. So I would compare the integrand to the function $x^{-3/2}.$