1. **Problem statement:** Determine whether the integral $$\int_1^\infty \frac{2 + \sin x}{\sqrt{x}} \, dx$$ converges or diverges using the comparison theorem. 2. **Recall the comparison theorem:** If $$0 \leq f(x) \leq g(x)$$ for all $$x$$ in the interval and $$\int_a^\infty g(x) \, dx$$ converges, then $$\int_a^\infty f(x) \, dx$$ also converges. Conversely, if $$\int_a^\infty g(x) \, dx$$ diverges and $$f(x) \geq g(x) \geq 0$$, then $$\int_a^\infty f(x) \, dx$$ diverges. 3. **Analyze the integrand:** Since $$\sin x$$ oscillates between $$-1$$ and $$1$$, we have $$1 \leq 2 + \sin x \leq 3$$ Therefore, $$\frac{1}{\sqrt{x}} \leq \frac{2 + \sin x}{\sqrt{x}} \leq \frac{3}{\sqrt{x}}$$ 4. **Compare with simpler integrals:** Consider the integral $$\int_1^\infty \frac{1}{\sqrt{x}} \, dx = \int_1^\infty x^{-1/2} \, dx$$ 5. **Evaluate the simpler integral:** $$\int_1^\infty x^{-1/2} \, dx = \lim_{t \to \infty} \int_1^t x^{-1/2} \, dx = \lim_{t \to \infty} \left[ 2x^{1/2} \right]_1^t = \lim_{t \to \infty} (2\sqrt{t} - 2) = \infty$$ This integral diverges. 6. **Conclusion:** Since $$\int_1^\infty \frac{1}{\sqrt{x}} \, dx$$ diverges and $$\frac{2 + \sin x}{\sqrt{x}} \geq \frac{1}{\sqrt{x}}$$, by the comparison theorem, the original integral $$\int_1^\infty \frac{2 + \sin x}{\sqrt{x}} \, dx$$ also diverges. --- **Final answer:** The integral $$\int_1^\infty \frac{2 + \sin x}{\sqrt{x}} \, dx$$ diverges.