The solution of $(1 + y^{2}) dx = ({Tan}^{- 1} y - x) dy$ is

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$x = ({Tan}^{- 1} y - 1) e^{{Tan}^{- 1} y} - y + c$

${xe}^{Tan - 1} y = ({Tan}^{- 1} t - 1) e^{{Tan}^{- 1}} - x + c$

$x = ({Tan}^{- 1} Y + 1) e^{{Tan}^{- 1}} - x + y + c$

${xe}^{{Tan}^{- 1}} = ({Tan}^{- 1} y - 1) e^{{Tan}^{- 1} y} + c$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{d x}{d y} = \frac{\tan^{- 1} y}{1 + y^{2}} - \frac{x}{1 + y^{2}} \\ \Rightarrow \frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{\tan^{- 1} y}{1 + y^{2}} \\ I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} \\ S o l u t i o n i s \\ x (I . F) = \int \frac{\tan^{- 1} y}{1 + y^{2}} e^{\tan^{- 1} y} d y \end{array}$

$\begin{array}{l} L e t \tan^{- 1} y = v \\ \frac{1}{1 + y^{2}} d y = d v \\ \Rightarrow x e^{\tan^{- 1} y} = \int v e^{v} \\ = v e^{v} - e^{v} + c \\ \Rightarrow x e^{\tan^{- 1} y} = T a n^{- 1} y e^{\tan^{- 1} y} - e^{\tan^{- 1} y} + c \end{array}$