Consider equation
\begin{equation}
M(x,y)\, dx + N(x,y)\, dy=0.
\label{A}
\end{equation}
It is exact iff
\begin{equation}
M_y -N_x=0.
\label{B}
\end{equation}
If (\ref{A}) is not exact (i. e. (\ref{B}) is violated we a looking for integrating factor $\mu(x,y)$ such that after multiplication by $\mu$ equation becomes exact:
\begin{equation}
(\mu M)_y -(\mu N)_x=0 \iff \mu_y M-\mu_x N + \mu (M_y-N_x)=0.
\label{C}
\end{equation}
Such factor is not unique and generally it is difficult to find. However
if $(M_y-N_x)/N=f(x)$ depends only on $x$ we look at integrating factor $\mu=\mu(x)$ and equation (\ref{C}) becomes
\begin{multline}
-\mu' N + \mu (M_y-N_x)=0 \iff (\log \mu)' = (M_y-N_x)/N=f(x) \implies\\ \log \mu (x)= \int f(x)\, dx.\qquad
\label{D}
\end{multline}
if $(M_y-N_x)/M=g(y)$ depends only on $y$ we look at integrating factor $\mu=\mu(y)$ and equation (\ref{C}) becomes
\begin{multline}
\mu' M + \mu (M_y-N_x)=0 \iff (\log \mu)' =- (M_y-N_x)/M=g(y) \implies\\
\log \mu(y) = -\int g(y)\, dy.\qquad
\label{E}
\end{multline}
if $(M_y-N_x)/(xM-yN)=h(xy)$ depends only $xy$ we look at integrating factor $\mu=\mu(xy)$ and equation (\ref{C}) becomes
\begin{multline}
\mu' (xM+yN) + \mu (M_y-N_x)=0 \iff (\log \mu)' =- (M_y-N_x)/(xM+yN)=\\
-h(xy) \implies
\log \mu (z) = -\int h(z)\, dz.\qquad
\label{F}
\end{multline}
After such factor is found, we can integrate an equation
\begin{equation}
\mu (x,y) M(x,y)\, dx + \mu(x,y) N(x,y)\, dy=0.
\label{G}
\end{equation}