Integrate the function: $\frac{1}{x-x^{3}}$

(D) We have $\frac{1}{x-x^{3}} = \frac{1}{x(1-x^{2})} = \frac{1}{x(1-x)(1+x)}$.
Using partial fractions,let $\frac{1}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} \dots (1)$.
Multiplying by $x(1-x)(1+x)$,we get $1 = A(1-x^{2}) + Bx(1+x) + Cx(1-x)$.
Setting $x=0$,we get $A=1$.
Setting $x=1$,we get $1 = B(1)(2) \Rightarrow B = \frac{1}{2}$.
Setting $x=-1$,we get $1 = C(-1)(2) \Rightarrow C = -\frac{1}{2}$.
Substituting these values into $(1)$,we have $\frac{1}{x-x^{3}} = \frac{1}{x} + \frac{1}{2(1-x)} - \frac{1}{2(1+x)}$.
Integrating both sides with respect to $x$:
$\int \frac{1}{x-x^{3}} dx = \int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{1-x} dx - \frac{1}{2} \int \frac{1}{1+x} dx$.
$= \log |x| - \frac{1}{2} \log |1-x| - \frac{1}{2} \log |1+x| + C$.
$= \log |x| - \frac{1}{2} \log |(1-x)(1+x)| + C$.
$= \log |x| - \frac{1}{2} \log |1-x^{2}| + C$.
$= \frac{1}{2} \log |x^{2}| - \frac{1}{2} \log |1-x^{2}| + C$.
$= \frac{1}{2} \log \left| \frac{x^{2}}{1-x^{2}} \right| + C$.