How do I find the singular solution of the differential equation $y' = \frac{y^2 + 1}{xy + y}$? I start out with the separable differential equation, 
$$y' =\frac{dy}{dx} = \frac{y^2 + 1}{xy + y} = \frac{y^2 + 1}{y(x+1)}$$
Thus, $\frac{1 }{x+1}dx = \frac{y }{y^2 + 1}dy$.
Then integrating both sides of the equation, I get 
$$\ln(x+1) = \frac{1}{2}\ln(y^2 +1) + C$$
Now, $e^{\ln(x+1)}$ = $e^{\frac{1}{2}\ln(y^2 +1) + C}$. So...
$$(x+1) = e^C(y^2 + 1)^{\frac{1}{2}}$$
I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.

I believe that $x + 1 = \pm e^C\sqrt{y^2 + 1}$ is the correct general solution. You can rewrite it as: $x + 1 = C\sqrt{y^2 + 1}$, where $C = \pm e^c \neq 0$.
As for your concern about the singular solution, I don't think the original equation has $x = -1$ as the singular solution because $x = -1$ makes the original equation undefined.
If only the original equation is initially given in the form:
\begin{align}y'(x + xy) = y^2 + 1\end{align}
then the solution $x = -1$ does not make the original equation undefined and therefore is singular.
Apologize for bad Engish.
