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

$\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right|+c$: where c is constant of integration.

The correct answer is Option (1) → $\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right|+c$: where c is constant of integration.

$\text{Integral to evaluate: }\displaystyle \int \frac{1}{x - x^5}\,dx$

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

$=\int \frac{1}{x(1-x^2)(1+x^2)}\,dx$

Rewrite:

$\frac{1}{x - x^5}=\frac{1}{x(1-x^4)} =\frac{1}{x}+\frac{x^3}{1-x^4}$

Integrate term-wise:

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

$\int \frac{x^3}{1-x^4}\,dx =\frac{1}{4}\ln|1-x^4|$

Combine:

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

Simplify:

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

The value of the integral is $\frac{1}{4}\ln\left|\frac{x^4}{1-x^4}\right|+C$.