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Answer
$$ \displaystyle\int \dfrac{{\mathrm{e}}^{{x}^{\frac{1}{3}}}{\cdot}{2}^{{\mathrm{e}}^{{x}^{\frac{1}{3}}}}}{{x}^{\frac{2}{3}}}\, \mathrm d x = {{3\,2^{e^{x^{{{1}\over{3}}}}}}\over{\ln 2}} $$
Explanation
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