Answer
$\dfrac{1}{3} (1+x^4)^{3/4}+c$
Work Step by Step
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$
where $c$ is a constant of proportionality.
Substitute $k=1+x^4 \implies 4x^3 dx =dk$
Thus,
$x^3(1+x^4)^{-1/4} dx=\dfrac{1}{4} \int(1+x^4)^{-1/4} (4x^3dx)=\dfrac{1}{4} \int k^{-1/4} dk$
or, $=\dfrac{1}{4} [\dfrac{k^{-1/4+1}}{-1/4+1}]+c$
or, $=\dfrac{1}{3} k^{3/4}+c$
plug back in: $k=1+x^4 $, we have
$=\dfrac{1}{3} (1+x^4)^{3/4}+c$
