# The solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1} y} )\frac{dy}{dx} = 0$ is :
( A ) $x e^{\tan^{-1} y} = \tan^{-1} y +k$
( B ) $(x-2) = ke^{-\tan^{-1} y}$
( C ) $x e^{2\tan^{-1} y} = e^{\tan^{-1} y} +k$
( D ) $2 x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} +k$