The value of $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to (where C is an arbitrary constant)

Given: $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$

Factor out $x$ from the numerator:

$(x^4 - x) = x(x^3 - 1)$

$(x^4 - x)^{1/4} = [x(x^3 - 1)]^{1/4} = x^{1/4} \cdot (x^3 - 1)^{1/4}$

The integral becomes:

$\int \frac{x^{1/4} \cdot (x^3 - 1)^{1/4}}{x^5} dx$

$= \int \frac{(x^3 - 1)^{1/4}}{x^{19/4}} dx$

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Rewrite the expression:

$\int \frac{(x^3 - 1)^{1/4}}{x^{19/4}} dx$

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

Factor out $x^{3/4}$ from $(x^3 - 1)^{1/4}$:

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

$= x^{3/4} \cdot \left(1 - \frac{1}{x^3}\right)^{1/4}$

The integral simplifies to:

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

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Let $u = 1 - \frac{1}{x^3}$

$u = 1 - x^{-3}$

$\frac{du}{dx} = 3x^{-4}$

$= \frac{3}{x^4}$

Therefore:

$\frac{1}{x^4}dx = \frac{1}{3}du$

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The integral becomes:

$\int u^{1/4} \cdot \frac{1}{3} du$

$= \frac{1}{3} \int u^{1/4} du$

$= \frac{1}{3} \cdot \frac{u^{5/4}}{5/4}$

$= \frac{1}{3} \cdot \frac{4}{5} \cdot u^{5/4}$

$= \frac{4}{15} u^{5/4} + C$

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Substitute back $u = 1 - \frac{1}{x^3}$:

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