If the general solution of $(1+y^2)dx = (\tan^{-1}y - x)dy$ is $x = f(y)+ce^{-\tan^{-1}y}$, then $f(y)=$
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To solve a first-order linear differential equation of the form $y' + P(x)y = Q(x)$, first calculate the integrating factor $I(x) = e^{\int P(x)dx}$. The solution is then given by $y \cdot I(x) = \int Q(x)I(x)dx + C$. Remember that the variables can be swapped, as in this problem where $x$ is a function of $y$.