\int {\frac{{dx}}{{\sqrt[4]{{2x^2  + x^4 }}}}}  = \left| \begin{array}{l}
 x = \frac{1}{t} \\ 
 dx =  - \frac{1}{{t^2 }}dt \\ 
 \end{array} \right| = \int {\frac{{ - \frac{1}{{t^2 }}dt}}{{\sqrt[4]{{\frac{2}{{t^2 }} + \frac{1}{{t^4 }}}}}}}  =  - \int {\frac{{\frac{1}{t}dt}}{{\sqrt[4]{{2t^2  + 1}}}}}  =  - \int {\frac{{dt}}{{t\sqrt[4]{{2t^2  + 1}}}}}  =  - \int {\frac{{tdt}}{{t^2 \sqrt[4]{{2t^2  + 1}}}}}  = \left| \begin{array}{l}
 2t^2  + 1 = u^4  \to t^2  = \frac{{u^4  - 1}}{2} \\ 
 tdt = u^3 du \\ 
 \end{array} \right| =  - \int {\frac{{u^3 du}}{{\frac{{u^4  - 1}}{2}u}}}  =  - \int {\frac{{2u^2 du}}{{u^4  - 1}}}  =  - 2\int {\left[ {\frac{1}{{4\left( {u - 1} \right)}} - \frac{1}{{4\left( {u + 1} \right)}} + \frac{1}{{2\left( {u^2  + 1} \right)}}} \right]du}  = ...