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

The integral to solve: $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$

Factor out $x$ from the numerator:

$(x^4 - x)^{1/4}$

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

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

Rewrite the integral:

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

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

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

This can be rewritten as $u = \frac{x^3 - 1}{x^3}$, which means $x^3 - 1 = x^3 u$

Find $du$:

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

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

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

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

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From $x^3 - 1 = x^3 u$:

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

Substitute into the integral:

$\int \frac{x^{3/4} \cdot u^{1/4}}{x^{19/4}} \cdot \frac{x^4}{3} du$

$= \int \frac{x^{3/4} \cdot u^{1/4} \cdot x^4}{x^{19/4} \cdot 3} du$

Combine powers of $x$ in the numerator:

$x^{3/4} \cdot x^4 = x^{19/4}$

$= \int \frac{x^{19/4} \cdot u^{1/4}}{x^{19/4} \cdot 3} du$

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

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

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

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

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

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

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