$\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} d x$ is equal to

$\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{\frac{5}{4}}+c$

$I=\int \frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5} dx$.  Put  $1-\frac{1}{x^3}=t$

∴   $\frac{3}{x^4} d x=d t$,     ∴  $I=\frac{1}{3} \int t^{\frac{1}{4}} d t$

$=\frac{1}{3} . \frac{t^{5 / 4}}{5 / 4}+c=\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5 / 4}+c$

Hence (1) is the correct answer.