Show that

$\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1-x)(1+x)}$

Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}$         ...(1)

$\Rightarrow 1=A\left(1-x^{2}\right)+B x(1+x)+C x(1-x)$

$\Rightarrow 1=A-A x^{2}+B x+B x^{2}+C x-C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$-A+B-C=0$

$B+C=0$

$A=1$

On solving these equations, we obtain

$A=1, B=\frac{1}{2}$, and $C=-\frac{1}{2}$

From equation (1), we obtain

$\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}$

$\Rightarrow \int \frac{1}{x(1-x)(1+x)} d x=\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{1-x} d x-\frac{1}{2} \int \frac{1}{1+x} d x$

$=\log |x|-\frac{1}{2} \log |(1-x)|-\frac{1}{2} \log |(1+x)|$

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

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

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

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