Answer
Diverges
Work Step by Step
$\displaystyle \int_{1}^{\infty}\frac{1+\sin^2x}{\sqrt x}\, dx$
Find a smaller function than $\displaystyle \frac{1+\sin^2x}{\sqrt x}$:
$\displaystyle \frac{1}{\sqrt{x}} \le \frac{1+\sin^2x}{\sqrt x}$ since $\displaystyle 0 \le \sin^2x \le 1$.
From the textbook, we know that
$\displaystyle \int_{1}^{\infty}\frac{1}{x^p}\, dx$ diverges when $\displaystyle p \le 1$.
Therefore, the smaller-valued function $\displaystyle \int_{1}^{\infty}\frac{1}{\sqrt x}\, dx$ diverges, and so does $\displaystyle \int_{1}^{\infty}\frac{1+\sin^2x}{\sqrt x}\, dx$ by the comparison theorem.
