Indefinite integral...(1)

Dividing numerator and denominator by $x^5$ ,we get, $\int(\dfrac{x^4}{x^{20}}-\dfrac{x}{x^{20}})^\frac{1}{4}dx$ $=\int(x^{-16}-x^{-19})^\frac{1}{4}dx$ $=\int x^{-4}(1-x^{-3})^{\frac{1}{4}}dx$ Now,substituting $1-x^{-3}=z$ we get, $x^{-4}dx=\dfrac{1}{3}*dz.$ The integral now becomes, $=\int\dfrac{1}{3}*z^{\frac{1}{4}}dz$ $=z^{\frac{5}{4}}*\dfrac{4}{5}*\dfrac{1}{3}$ $=\dfrac{4}{15}*(1-x^{-3})^{\frac{5}{4}}.$

$\begin{aligned} I & = \int \frac {(x^4-x)^\frac 14}{x^5} dx \\ & = \int \left(\frac {x^4}{x^{20}} -\frac x{x^{20}} \right)^\frac 14 dx \\ & = \int \left(x^{-16} - x^{-19} \right)^\frac 14 dx \\ & = \int \color{#3D99F6}{x^{-4} \left(1- x^{-3} \right)^\frac 14} dx \end{aligned}$

Now, note that $\dfrac d{dx} \left(1- x^{-3} \right)^\frac 54 = \dfrac 54 \left(1- x^{-3} \right)^\frac 14 \cdot 3x^{-4} = \dfrac {15}4 \color{#3D99F6}{x^{-4} \left(1- x^{-3} \right)^\frac 14}$ .

$\implies I = \boxed{\dfrac 4{15} \left(1- \dfrac 1{x^3} \right)^\frac 54 + C}$