Is there any exact formula or at least exact inequalities for the following intehral

$$ \int_{2}^{x}{{\rm d}t \over \left\lfloor\vphantom{\large h}% \log(x)/\log(t)\right\rfloor \log\left(t\right)} $$

where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.

added:

When I use

$$ x-1<\lfloor x\rfloor\le x $$ I get $$ \frac{x-2}{\log (x)}=\int_2^x\frac{{\mathrm d}t}{\log (x)}\leq \int_2^x\frac{{\mathrm d}t}{\left\lfloor\vphantom{\large h}\log (x)/\log (t)\right\rfloor\log (t)}\le \int_2^x\frac{\mathrm dt}{\log (x)-\log (t)} $$ but they are not exact enough. I need more closer bounds.