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∫01({1x}{2x})dxx \displaystyle \int_{0}^{1}{\left(\left\{ \frac{1}{x}\right\} \{ 2x \}\right) \frac{dx}{x} }∫01({x1}{2x})xdx
The above integral is equal to A+lnB−Cγ,A+\ln{B}-C\gamma,A+lnB−Cγ, where A,B,CA,B,CA,B,C are all positive integers.
Find A+B+C.A+B+C.A+B+C.
Note:{} \{ \} {} denotes the fractional part and γ\gammaγ denotes the Euler-Mascheroni constant.
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