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

$\frac{1}{5}\log_e\left|\frac{x^5-1}{x^5}\right|+C$: C is an arbitrary constant

The correct answer is Option (1) → $\frac{1}{5}\log_e\left|\frac{x^5-1}{x^5}\right|+C$: C is an arbitrary constant

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

Use substitution: Let $t = x^5 \Rightarrow dt = 5x^4 \, dx$

So, $dx = \frac{dt}{5x^4}$

Substitute into the integral:

$\int \frac{1}{x(x^5 - 1)} \, dx = \int \frac{1}{x(t - 1)} \cdot \frac{dt}{5x^4}$

$= \int \frac{1}{5x^5(t - 1)} \, dt$

Since $x^5 = t$, so $x^5(t - 1) = t(t - 1)$

Thus, the integral becomes:

$\int \frac{1}{5t(t - 1)} \, dt$

Use partial fractions: $\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}$

So, the integral becomes:

$\frac{1}{5} \int \left( \frac{1}{t - 1} - \frac{1}{t} \right) dt$

$= \frac{1}{5} \left[ \ln |t - 1| - \ln |t| \right] + C$

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

Substitute $t = x^5$:

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