The value of ${\int }_0^{\left[x\right]} \frac{2^t}{2^{\left[t\right]}}dt$(where [ ] denotes greatest integer function)

$I={\int }_0^{\left[x\right]} \frac{2^t}{2^{\left[t\right]}}dt={\int }_0^{\left[x\right]} 2^{t-\left[t\right]}dt$

The function $g\left(t\right)=2^{t-\left[t\right]}$ is periodic with period 1, 

Therefore

$\begin{array}{l}I=\left[x\right]{\int }_0^1 2^{t-\left[t\right]}dt=\left[x\right]{\int }_0^1 2^{t}dt=\frac{\left[x\right]}{\log 2}[2t{]}_0^1\\ =\frac{\left[x\right]}{\log 2}\end{array}$