For $x \in \mathbb{R} - \{-1,0,1\}$, $\int \frac{1}{x - x^5}dx$ is equal to

Look at the denominator: $x - x^5$

Factor out $x$:

$x - x^5 = x(1 - x^4)$

The integral becomes:

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

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Using partial fractions:

$\frac{1}{x(1-x^4)} = \frac{A}{x} + \frac{f(x)}{1-x^4}$

Multiply both sides by $x(1-x^4)$:

$1 = A(1-x^4) + x \cdot f(x)$

Put $x = 0$:

$1 = A(1) + 0$

$A = 1$

Substitute $A=1$:

$1 = (1-x^4) + x \cdot f(x)$

$x \cdot f(x) = x^4$

$f(x) = x^3$

Therefore:

$\frac{1}{x(1-x^4)} = \frac{1}{x} + \frac{x^3}{1-x^4}$

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

First integral:

$\int \frac{1}{x}dx = \ln|x|$

For the second integral, let $u = 1-x^4$

Then $\frac{du}{dx} = -4x^3$, so $x^3dx = -\frac{1}{4}du$

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

$= -\frac{1}{4}\ln|u|$

$= -\frac{1}{4}\ln|1-x^4|$

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$\int \frac{1}{x - x^5}dx = \ln|x| - \frac{1}{4}\ln|1-x^4| + c$

$= \frac{1}{4}\left[4\ln|x| - \ln|1-x^4|\right] + c$

$= \frac{1}{4}\left[\ln|x^4| - \ln|1-x^4|\right] + c$

$= \frac{1}{4}\ln\left|\frac{x^4}{1-x^4}\right| + c$

Therefore, the answer is:

$\boxed{\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right| + c}$

Note: $\log_e$ is the same as $\ln$ (natural logarithm)