Prove

Let $x^{4}=t$

$\therefore 4 x^{3} d x=d t$

$\Rightarrow \int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} d x=\frac{1}{4} \int \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} d t$    ...(1)

Let $\tan ^{-1} t=u$

$\therefore \frac{1}{1+t^{2}} d t=d u$

From (1), we obtain

$\begin{aligned} \int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right) d x}{1+x^{8}} &=\frac{1}{4} \int \sin u d u \\ &=\frac{1}{4}(-\cos u)+\mathrm{C} \end{aligned}$

$=\frac{-1}{4} \cos \left(\tan ^{-1} t\right)+\mathrm{C}$

$=\frac{-1}{4} \cos \left(\tan ^{-1} x^{4}\right)+\mathrm{C}$