The integrating factor of differential equation
$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - xy = 1$ is

Given that,

$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - xy = 1$

$\Rightarrow$ $\frac{{dy}}{{dx}} - \frac{x}{{1 - {x^2}}}y = \frac{1}{{1 - {x^2}}}$

which is a linear differential equation.
$\therefore$ ${\rm{IF}} = {e^{ - \int {\frac{x}{{1 - {x^2}}}} dx}}$

Put $1 - {x^2} = t \Rightarrow - 2xdx = dt \Rightarrow xdx = - \frac{{dt}}{2}$

Now, ${\rm{IF}} = {e^{\frac{1}{2}\int {\frac{{{\rm{Jt}}}}{t}} }} = {e^{\frac{1}{2}\log t}} = {e^{\frac{1}{2}\log \left( {1 - {x^2}} \right)}} = \sqrt {1 - {x^2}}$