If $\int \frac{dx}{x-x^3}=A\ln \left|\frac{x^2}{1-x^2}\right|+c$, then the value of $A$ is:

Let $I=\int \frac{1}{x(1-x^2)}dx$.Using partial fractions: $\frac{1}{x(1-x^2)}=\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1/2}{1-x}-\frac{1/2}{1+x}$.$I=\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{1-x}dx-\frac{1}{2}\int \frac{1}{1+x}dx$$I=\ln |x|-\frac{1}{2}\ln |1-x|-\frac{1}{2}\ln |1+x|+c$$I=\frac{1}{2}[2\ln |x|-(\ln |1-x|+\ln |1+x|)]+c$$I=\frac{1}{2}[\ln |x^2|-\ln |1-x^2|]+c$$I=\frac{1}{2}\ln \left|\frac{x^2}{1-x^2}\right|+c$.Comparing with the given form, $A=\frac{1}{2}$.