Integral of a floor function.
Observe that when we only substitute $\frac{1}{x} = y$
$$I = \int_1^\infty \frac{dy}{\lfloor y\rfloor y^2} = \sum_{n=1}^\infty \frac{1}{n}\int_n^{n+1}\frac{dy}{y^2}$$
$$ = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+1} = \frac{\pi^2}{6} - 1$$