If I have an exact DE of the form
$M(x,y) dx + N(x,y) dy = 0$,
then we solve it using the following steps:
Check exactness: $\dfrac{\partial{M}}{\partial{y}} = \dfrac{\partial{N}}{\partial{x}}$? If yes, then proceed.
Assume that $\dfrac{\partial{f}}{\partial{x}} = M(x,y)$ and $\dfrac{\partial{f}}{\partial{y}} = N(x,y)$.
Integrate $M(x,y)$ with respect to $x$ and add an arbitrary function of $y$, $g(y)$, as the integration constant. This gives us $f$.
Differentiate the result from step 3 with respect to $y$, which gives $\dfrac{\partial{f}}{\partial{y}}$, and compare with $N(x,y)$. Solve this for $g'(y)$.
Integrate $g'(y)$ with respect to $y$ to find an expression for $g(y)$. Do not add an integration constant.
Substitute $g(y)$ into the result obtained in step 3 to obtain $f(x,y)$.
The general solution is $f(x,y) = C$, where $C$ is our constant of integration.
A similar procedure can be stated so that in step 3, $N(x,y)$ is integrated with respect to $y$.
My question is, why do we not add an integration constant at step 5?
I would greatly appreciate it if people could please take the time to clarify this.