Find the general solution of the DE : dy/dx = (1+y^2)/(1+x^2)

\(\frac{dy}{dx}\) = \(\frac{1+y^2}{1+x^2}\)

dx . 1+x² = dy . 1+y²  and then integrate to get : 

\(\tan^{-1} {x}\) = \(\tan^{-1} {y}\) + C , or

\(\tan^{-1} {y}\) = \(\tan^{-1} {x}\) + C, or

\(\tan^{-1} {x}\) - \(\tan^{-1} {y}\) = C

Note : -ve sign can be incorporated into the constant C