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

Given differential equation is
$\left( {{x^2} - 1} \right)\frac{{dy}}{{dx}} + 2xy = \frac{1}{{{x^2} - 1}}$

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

which is a linear differential equation.
On comparing it with

$\frac{{dy}}{{dx}} + Py = Q,$ we get

$P = \frac{{2x}}{{{x^2} - 1}},$ $Q = \frac{1}{{{{\left( {{x^2} - 1} \right)}^2}}}$

$IF = {e^{\int P dx}} = {e^{\int {{{\left( {\frac{{2x}}{{{x^2} - 1}}} \right)}^{dx}}} }}$

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

$\therefore$ ${\rm{IF}} = {e^{\int {\frac{{dt}}{t}} }} = {e^{\log t}} = t = \left( {{x^2} - 1} \right)$

The complete solution is

$y \cdot {\rm{IF}} = \int Q \cdot {\rm{IF + K}}$

$\Rightarrow$ $y \cdot \left( {{x^2} - 1} \right) = \int {\frac{1}{{{{\left( {{x^2} - 1} \right)}^2}}}} \cdot \left( {{x^2} - 1} \right)dx + K$

$\Rightarrow$ $y \cdot \left( {{x^2} - 1} \right) = \int {\frac{{dx}}{{\left( {{x^2} - 1} \right)}}} + K$

$\Rightarrow$ $y \cdot \left( {{x^2} - 1} \right) = \frac{1}{2}\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + K$