Explanation

(a)
Given, $\left( x^{2} + 1 \right)\frac{dy}{dx} + 2xy = x^{2} - 1$
or $\frac{dy}{dx} + \frac{2x}{1 + x^{2}}y = \frac{x^{2} - 1}{x^{2} + 1}$
It is a linear differential equation.
On comparing with the standard equation $\frac{dy}{dx} + Py = Q$, we get
$P = \frac{2x}{1 + x^{2}},Q = \frac{x^{2} - 1}{x^{2} + 1}$
$\therefore IF = e^{\int_{}^{}{P\ dx}} = e^{\int_{}^{}{\frac{2x}{1 + x^{2}}dx}}$
$= e^{\log{(1 + x^{2})}} = 1 + x^{2}$