. $\cfrac{{{x^3}}}{{\sqrt {1 - {x^8}} }}$

Let $I = \int {\cfrac{{{x^3}}}{{\sqrt {1 - {x^8}} }}} dx = \cfrac{1}{4}\int {\cfrac{{4{x^3}}}{{\sqrt {1 - {{\left( {{x^4}} \right)}^2}} }}} dx$

Put ${x^4} = t$ $\Rightarrow$

$4{x^3}dx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\cfrac{{dt}}{{\sqrt {1 - {t^2}} }} = \cfrac{1}{4}{{\sin }^{ - 1}}\left( t \right) + C = \cfrac{1}{4}{{\sin }^{ - 1}}\left( {{x^4}} \right) + C}$