integrate

\frac{[x]}{3ln2} ?

not too sure of it though

ok

2^{x} has a period 1 so 2^{3x} has a period 1/3

and from propety of DI \int_{0}^{nT}{f(x)dx }=n\int_{0}^{T}{f(x)dx }

here T=1/3 n=[x]

so \int_{0}^{[x]/3}{2^{\left\{3x \right\}}dx }=[x]\int_{0}^{1/3}{2^{\left\{3x \right\}}dx }\\

now since x lies from 0 tp 1/3 so 3x lies from 0 to 1 so {3x}=3x

hence
[x]\int_{0}^{1/3}{2^{\left\{3x \right\}}dx }=[x]\int_{0}^{1/3}{2^{3x}dx }\\

put 3x=t

\frac{[x]}{3}\int_{0}^{1}{2^{t}dx }=\frac{[x]}{3ln2}

yeah seems correct... wonder why i dint think of the periodicity