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

When we see a product in the denominator like $x(x^5-1)$, partial fractions can be applied.

$\frac{1}{x(x^5-1)} = \frac{A}{x} + \frac{P(x)}{x^5-1}$

Since $(x^5-1)$ is degree 5, the numerator of the second fraction must be a polynomial of degree at most 4.

---

Multiply both sides by $x(x^5-1)$:

$1 = A(x^5-1) + x \cdot P(x)$

Put $x = 0$:

$1 = A(0-1) = -A$

$A = -1$

Substitute $A = -1$:

$1 = -1(x^5-1) + xP(x)$

$1 = -x^5 + 1 + xP(x)$

$x^5 = xP(x)$

$P(x) = x^4$

The partial fraction decomposition:

$\frac{1}{x(x^5-1)} = \frac{-1}{x} + \frac{x^4}{x^5-1}$

---

$\int \frac{1}{x(x^5-1)} dx = \int \frac{-1}{x} dx + \int \frac{x^4}{x^5-1} dx$

First integral:

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

Second integral, the derivative of $(x^5-1)$ is $5x^4$.

Let $u = x^5-1$, then $du = 5x^4 dx$, which means $x^4 dx = \frac{1}{5}du$

$\int \frac{x^4}{x^5-1} dx = \int \frac{1}{u} \cdot \frac{1}{5} du$

$= \frac{1}{5}\ln|u|$

$= \frac{1}{5}\ln|x^5-1|$

---

$\int \frac{1}{x(x^5-1)} dx = -\ln|x| + \frac{1}{5}\ln|x^5-1| + C$

$= \frac{1}{5}\ln|x^5-1| - \frac{1}{5} \cdot 5\ln|x| + C$

$= \frac{1}{5}\left(\ln|x^5-1| - 5\ln|x|\right) + C$

$= \frac{1}{5}\left(\ln|x^5-1| - \ln|x^5|\right) + C$

$= \frac{1}{5}\ln\left|\frac{x^5-1}{x^5}\right| + C$

Therefore, the answer is $\frac{1}{5}\log_e \left|\frac{x^5-1}{x^5}\right| + C$