The value of $\int\limits_0^{[x]} \frac{2^x}{2^{[x]}} d x$, is

$\frac{[x]}{\log 2}$

We have,

$\int\limits_0^{[x]} \frac{2^x}{2^{[x]}} d x=\int\limits_0^{[x]} 2^{x-[x]} d x$

Since, $2^{x-[x]}$ is a periodic function with period one unit.

∴  $\int\limits_0^{[x]} 2^{x-[x]} d x=[x] \int\limits_0^1 2^{x-[x]} d x=[x] \int\limits_0^1 2^x d x$

$\Rightarrow \int\limits_0^{[x]} 2^{x-[x]} d x=\frac{[x]}{\log 2}\left[2^x\right]_0^1=\frac{[x]}{\log 2}(2-1)=\frac{[x]}{\log 2}$