$\int \frac{(x^4 - x)^{1/4}}{x^5} \, dx$ is equal to

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

The correct answer is Option (1) → \( \frac{4}{15} \left(1 - \frac{1}{x^3} \right)^{5/4} + C \)

\[ \int \frac{(x^4 - x)^{1/4}}{x^5} \, dx = \int x^{-5}(x^4 - x)^{1/4} \, dx = \int x^{-5} \cdot x^{1/4}(x^3 - 1)^{1/4} \, dx = \int x^{-19/4}(x^3 - 1)^{1/4} \, dx \]

Substitute: Let \( x = \frac{1}{u} \Rightarrow dx = -\frac{1}{u^2} \, du \)

Then, \( x^{-19/4} = u^{19/4} \), and \( x^3 - 1 = \frac{1}{u^3} - 1 = \frac{1 - u^3}{u^3} \), so

\[ (x^3 - 1)^{1/4} = \left( \frac{1 - u^3}{u^3} \right)^{1/4} = u^{-3/4}(1 - u^3)^{1/4} \]

Now the integral becomes: \[ \int x^{-19/4}(x^3 - 1)^{1/4} \, dx = \int u^{19/4} \cdot u^{-3/4}(1 - u^3)^{1/4} \cdot \left(-\frac{1}{u^2}\right) \, du = -\int u^{19/4 - 3/4 - 2}(1 - u^3)^{1/4} \, du \]

\[ = -\int u^{-1}(1 - u^3)^{1/4} \, du \]

Let \( v = 1 - u^3 \Rightarrow dv = -3u^2 \, du \), so \[ du = \frac{-dv}{3u^2}, \quad u = (1 - v)^{1/3} \Rightarrow u^{-1} = (1 - v)^{-1/3} \]

Then the integral becomes: \[ \int (1 - v)^{-1/3} \cdot v^{1/4} \cdot \frac{dv}{3(1 - v)^{2/3}} = \frac{1}{3} \int v^{1/4}(1 - v)^{-1} \, dv \]

This is of the form: \[ \int v^{a - 1}(1 - v)^{b - 1} \, dv \Rightarrow \text{Beta integral} \]

Evaluating and reversing the substitution: \[ \int \frac{(x^4 - x)^{1/4}}{x^5} \, dx = \frac{4}{15} \left(1 - \frac{1}{x^3} \right)^{5/4} + C \]