The differential equation: $ \arctan (y) = \arctan(x)+C .$ I solved the equation and stalled. Help with decision please.
$$(1+y^2)\,dx=(1+y^2)\,dy \iff \int \frac{dx}{1+x^2} = \int\frac{dy}{1+y^2}  $$ 
Transformed expression for the table of integrals.
$$ \arctan (y) = \arctan(x)+C $$
Prompt how to further transform expression.(Find the general solution)

The differential set
$$
(1+y^{2}) \,dx  = (1+x^{2})\, dy 
$$
can be integrated as seen by
\begin{align}
\int \frac{dy}{1+y^{2}} = \int \frac{dx}{1+x^{2}}
\end{align}
and leads to
\begin{align}
\tan^{-1}(y) = \tan^{-1}(x) + c
\end{align}
or
\begin{align}
y = \tan(\tan^{-1}(x) + c).
\end{align}
Now using 
\begin{align}
\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)}
\end{align}
the general result becomes
\begin{align}
y(x) = \frac{x+c_{1}}{1 - c_{1} x}
\end{align}
since $\tan(\tan^{-1}(x)) = x$ and $c_{1} = \tan(c)$ is a constant. 
