$\int \frac{\left(x-x^5\right)^{1 / 5}}{x^6} d x$ is equal to

$-\frac{5}{24}\left(\frac{1}{x^4}-1\right)^{6 / 5}+C$

Let

$I =\int \frac{\left(x-x^5\right)^{1 / 5}}{x^6} d x=\int\left(\frac{1}{x^4}-1\right)^{1 / 5} \frac{1}{x^5} d x$

$\Rightarrow I =-\frac{1}{4} \int\left(\frac{1}{x^4}-1\right)^{1 / 5}\left(\frac{-4}{x^5}\right) d x=-\frac{1}{4} \int\left(\frac{1}{x^4}-1\right)^{1 / 5} d\left(\frac{1}{x^4}-1\right)$

$\Rightarrow I =-\frac{5}{24}\left(\frac{1}{x^4}-1\right)^{6 / 5}+C$