Solution
= Solution
For $x\geq1$, $|\sin(1/\sqrt x)|\leq1/\sqrt x$, so
$$
|g(x)|\leq x^{-3/2},\qquad\int_1^\infty x^{-3/2}dx=2.
$$
Thus \b[$g$ is integrable] on this <interval>. But for $x>1$, $\sqrt{x^2-1}<x$, hence $h(x)>1/x$. The integral of $1/x$ diverges at infinity, so \b[$h$ is not integrable]. Defining $h(1)=0$ changes only a <null set> and has no effect on this conclusion.