$\cfrac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}$

: Let $I = \int {\cfrac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}} dx$

Put ${\tan ^{ - 1}}{x^4} = t$
$\Rightarrow$ $\cfrac{1}{{1 + {x^8}}} \cdot 4{x^3}dx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\sin tdt} = \cfrac{1}{4}\left( { - \cos t} \right) + C = - \cfrac{1}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + C$