$\int\limits_0^x \frac{2^t}{2^{[t]}} d t$, where [.] denotes the greatest integer function, and x ∈ R⁺, is equal to

$\frac{1}{\ln 2}\left([x]+2^{\{x\}}-1\right)$

Let $n \leq x<n+1$ where $n \in I, I \geq 0$

$I=\int\limits_0^X \frac{2^t}{2^{[t]}} d t=\int\limits_0^n 2^{\{t\}} d t+\int\limits_n^x 2^{\{t\}} d t$

$=n \int\limits_0^1 2^{\{t\}} d t+\int\limits_n^x 2^{\{t\}} d t=n \int\limits_0^1 2^t d t+\int\limits_0^x 2^{t-n} d t$

$=\left.n . \frac{2^t}{\ln 2}\right|_0 ^1+\left.\frac{1}{2^n} . \frac{2^t}{\ln 2}\right|_n ^x$

$=n . \frac{1}{\ln 2}(2-1)+\frac{1}{2^n . \ln 2}\left(2^x-2^n\right)=\frac{[x]+2^{\{x\}}-1}{\ln 2}$