$\int \frac{d x}{x-x^3}=A \log \left|\frac{x^2}{1-x^2}\right|+c$ then A is equal to

$\frac{1}{2}$

$I=\int \frac{d x}{x-x^3}=\int\frac{x-3}{x^{-2}-1}$

let $z=x^{-2}-1$

so $dz=-2x^{-3}dx⇒\frac{-1}{2}dz=x^{-3}dx$

so $I=\frac{-1}{2}\int\frac{dz}{z}=\frac{-1}{2}\log|z|+C$

$=\frac{1}{2}\log\left|\frac{1}{z}\right|+C=\frac{1}{2}\log\left|\frac{1}{x^{-2}-1}\right|+C$

$=\frac{1}{2}\log\left|\frac{x^2}{-x^2+1}\right|+C$

so $A=\frac{1}{2}$