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Answer
$$ \int {2}\pi{x}^{{\frac{{1}}{{3}}}}{\left(\sqrt{{{1}+\frac{{1}}{{{9}{x}^{{\frac{{4}}{{3}}}}}}}}\right)} \, d\,x = {{\pi\,\left(x^{{{1}\over{3}}}\,\ln \left({{\sqrt{9\,x^{{{4}\over{ 3}}}+1}+3\,x^{{{2}\over{3}}}}\over{3\,x^{{{2}\over{3}}}}}\right)-x^{{{1}\over{3}}}\,\ln \left({{\sqrt{9\,x^{{{4}\over{3}}}+1}-3\,x^{{{2 }\over{3}}}}\over{3\,x^{{{2}\over{3}}}}}\right)+6\,x\,\sqrt{9\,x^{{{ 4}\over{3}}}+1}\right)}\over{12\,x^{{{1}\over{3}}}}} $$
Explanation
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