## Calculus (3rd Edition)

The integral $\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1}$ diverges.
We are given the function $f(x)=\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1}$ Since, $\sqrt x \geq \sqrt x-1$, for $x \gt 1$, we have: $\dfrac{1}{\sqrt x} \leq \dfrac{1}{\sqrt x-1}$ The integral $\int_{1}^{\infty} \dfrac{d x}{\sqrt x}=\int_{1}^{\infty} \dfrac{d x}{x^{1/2}}$ diverges because $\dfrac{1}{2} \lt 1$ Therefore, by the comparison test, the integral $\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1}$ diverges.