If $Mdx+Ndy=0$ be an exact differential equation, then solution to that differential equation is/are

Question

If $Mdx+Ndy=0$ be a exact differential equation, then solution to that differential equation is/are

(1) $\int Mdx +\int Ndy-\int( \frac{\partial}{\partial y}\int Mdx)dy=c$

(2) $\int Mdx +\int Ndy-\int( \frac{\partial}{\partial x}\int N dy)dx=c$

(3) $\int Mdx +\int Ndy =c$, $N$ is free from $x$

(4) $\int Mdx +\int Ndy=c$, $M$ is free from $y$

My efforts

Differential equation and multi variable calculus are my weakest topics and I picked this question to fill the holes. I really need the help.

So let's come to question.

1. This equation is exact so I have $M_y=N_x$.

2. We say a function $f(x,y)$ satisfies an exact differential equation if $f_x=M$ and $f_y=N$

So what is the next step,

Let's take option, say $c$

$f(x,y)=\int Mdx +\int Ndy -c$,

Let us calculate $f_x$, that would be $\frac{\partial}{\partial x}(\int Mdx +\int Ndy -c)$

$$f_x=\frac{\partial}{\partial x}(\int Mdx) + \frac{\partial}{\partial x}(\int Ndy)-\frac{\partial}{\partial x}(c)$$

How do these switching of integral work. Can someone explain in detail by writing the rules. I will go through the theory on my own. Please help in all other option as well.

Edits

I did some Google search, I can use the Leibniz Integration rule here which says that

$$\frac{d}{dx}\int_{a}^{b} f(x, t)dt=\int\frac{\partial f(x, t)}{\partial x} dt$$

Using this we get $f_x=M$, right? $N$ is free from $x$ do $N_x=0$ and $\int M_xdx=M$ \

Edit 2

I think I have worked it on my own. It will take so much time to write everything here. I think all options are correct. Can someone please check it and confirm.