JEE Math Practice Question

The solution of the differential equation $$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is :

1. A.$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$$
2. B.$$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$$
3. C.$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
4. D.$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$$