$\cfrac{1}{{{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}$

Let $I = \int {\cfrac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} = \int {\cfrac{{dx}}{{{x^5}{{\left( {1 + \cfrac{1}{{{x^4}}}} \right)}^{3/4}}}}}$

Put $1 + \cfrac{1}{{{x^4}}} = t$ $\Rightarrow$ $- \cfrac{4}{{{x^5}}}dx = dt$ $\Rightarrow$ $\cfrac{1}{{{x^5}}}dx = - \cfrac{1}{4}dt$

$\therefore$ $I = - \cfrac{1}{4}\int {\cfrac{{dt}}{{{t^{3/4}}}}} = - \cfrac{1}{4}\int {{t^{ - 3/4}}} dt$

$= - \cfrac{1}{4}\cfrac{{{t^{1/4}}}}{{1/4}} + C = - {\left( t \right)^{1/4}} + C = - {\left( {1 + \cfrac{1}{{{x^4}}}} \right)^{1/4}} + C$