Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

Let $x^{4} = t$.
Then,$4x^{3} dx = dt$,which implies $x^{3} dx = \frac{1}{4} dt$.
Substituting this into the integral,we get:
$\int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} dx = \frac{1}{4} \int \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} dt$.
Now,let $\tan ^{-1} t = u$.
Then,$\frac{1}{1+t^{2}} dt = du$.
Substituting $u$ into the integral:
$\frac{1}{4} \int \sin u du = \frac{1}{4} (-\cos u) + C = -\frac{1}{4} \cos u + C$.
Substituting back $u = \tan ^{-1} t$ and $t = x^{4}$:
$= -\frac{1}{4} \cos \left(\tan ^{-1} x^{4}\right) + C$,where $C$ is the constant of integration.