Answer
$$ = {{8\,\sqrt{x}+3\,\sqrt{2}\,\ln \left(x^{{{1}\over{3}}}+\sqrt{2}\,x ^{{{1}\over{6}}}+1\right)-3\,\sqrt{2}\,\ln \left(x^{{{1}\over{3}}}- \sqrt{2}\,x^{{{1}\over{6}}}+1\right)-6\sqrt{2}\,\arctan \left({{2\,x^{{{1}\over{6}}}+\sqrt{2}}\over{\sqrt{2}}}\right)-6\sqrt{2}\,\arctan \left({{2\,x^{{{1}\over{6}}}-\sqrt{2}}\over{ \sqrt{2}}}\right)}\over{4}} $$
Explanation
This page was created using
Integral Calculator