This question somewhat confuses me with the function being defined as $1/x-1/x$,for $(0,1]$, and 0 where x=0. I'm also somewhat thrown off about . meaning integer part. Anyway, here is my attempt, which I am not sure if is correct:
Attempt:
The function is continuous at every point on [0,1]. except at x=0.
Lebesgue's Integrability Criterion states that a bounded function $f:[a,b]$ to $R$ is Riemann Integrable if and only if it is continuous almost everywhere on [a,b].
The function given is bounded on the interval [0,1]. and the point of discontinuity, x=0, is a null set, where: 
.
Thus, the function is continuous almost everywhere on [0,1]. and hence is Riemann integrable.
Updated Answer: The function is discontinuous at 0 and points of the form $1/n$, where n is a natural number, starting from 2,3....... (I believe the function is continuous at 1, that's why we start from 2). This set is clearly countable and thus forms a null set (the set of discontinuities forms a null set).
Thus, by Lebesgue's Integrability Criterion, the function is Riemann integrable on [0,1]. I believe the function is bounded by 1 from above and 0 from below.