$(1+x^2)\frac{dy}{dx}$ + y = $\tan^{-1}$ x $\frac{dy}{dx}$ + $\frac{y}{1+x^2}$ = $\frac{\tan^{-1}x}{1+x^2}$ ... (i) Given equation is linear with So, I.F. = $e^{\int \frac{1}{1+x^2}dx}$ = $e^{\tan^{-1}x}$ Solution of (i) $y e^{\tan^{-1}x}$ = ∫ $e^{\tan^{-1}x} \left( \frac{\tan^{-1}x}{1+x^2} \right)$ dx ... (ii) For R.H.S,let $\tan^{-1}$ x = t ⇒ $\frac{1}{1+tx^2}$ dx = dt By substituting in equation(ii) $y e^{\tan^{-1}x}$ = ∫ $e^{t}$ . tdt ⇒ $y e^{\tan^{-1}x}$ = $\left[ t e^{t} - e^{t} \right]$ + C ⇒ $y e^{\tan^{-1}x}$ = $e^{\tan^{-1}x}$ $(\tan^{-1}x - 1)$ + C ⇒ y = $\tan^{-1}x - 1 + C e^{-\tan^{-1}x}$