$\int \frac{1}{x-x^3}dx$ is equal to:

Let $I=\int \frac{1}{x-x^3}dx$ It can be written as: $I=\int \frac{xdx}{x^2\left(1-x^2\right)}$ (multiply and divide by x)

Put ${\mathrm{x}}^2=\mathrm{t}\Rightarrow 2\mathrm{xdx}=\mathrm{dt}\Rightarrow \mathrm{xdx}=\frac{\mathrm{dt}}{2}$ So, $\mathrm{I}=\int \frac{\mathrm{dt}}{2\mathrm{t}(1-\mathrm{t})}=\frac{1}{2}\int \frac{\mathrm{d}t}{\mathrm{t}(1-\mathrm{t})}$

$=\frac{1}{2}\int \left[\frac{1}{t}+\frac{1}{1-t}\right]dt$

$=\frac{1}{2}[\log t-\log (1-t\left)\right]+c$

Thus, $I=\frac{1}{2}\left[\log \frac{t}{1-t}\right]+c$

$=\frac{1}{2}\left[\log \frac{x^2}{1-x^2}\right]+c$