The general solution of differential equation $(\tan^{-1}y - x)dy = (1 + y^2)dx$ is:

Let $x$ be a function of $y.$ $\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2}$ $\Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y),$ where $P(y) = \frac{1}{1 + y^2}.$ Integrating factor, $\mu = e^{\int P(y) dy} = e^{\tan^{-1}y}.$ $\Rightarrow \frac{d}{dy}(xe^{\tan^{-1}y}) = e^{\tan^{-1}y} \frac{\tan^{-1}y}{1 + y^2}$ Let $t = \tan^{-1}y \Rightarrow dt = \frac{dy}{1 + y^2}$ $xe^{t} = \int t e^{t} dt = e^{t}(t - 1) + C$ $\Rightarrow x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$