${\displaystyle \int \frac{{(x-x^3)}^{1/3}}{x^4}dx}$

${\displaystyle \int \frac{x{\left[\frac{1}{x^2}-1\right]}^{\frac{1}{3}}}{x^4}}dx=$ $-\frac{1}{2}{\displaystyle \int t^{\frac{1}{3}}}dt$

$\frac{1}{x^2}-1=t\Rightarrow -\frac{2}{x^3}dx=dt$

                      $=-\frac{1}{2}\frac{t^{\frac{4}{3}}}{\left(\frac{4}{3}\right)}+C$

                      $=-\frac{3}{8}{\left(\frac{1}{x^2}-1\right)}^{\frac{4}{3}}+C$