Integration factor - First Order Nonlinear ODE I can't seem to find the proper integrating factor for this nonlinear first order ODE. I have even tried pulling a bunch of substitution and equation-manipulating tricks, but I can't seem to get a proper integrating factor.
$$\frac{1}{x}dx + \left(1+x^2y^2\right)dy = 0$$
EDIT: Due to MSE users complaining about my lack of proof of work, intent of conceptual understanding, etc, here is exactly why I am stuck.
To start off, this ODE is obviously inexact:
$$\frac{\partial}{\partial y}\left(\frac{1}{x}\right) \neq \frac{\partial}{\partial x}\left(1+x^2y^2\right)$$
And so in order to make this exact (if we choose to go down this route) we must (I'll stick to standard convention/notation) find a function $\mu$ such that if we multiply the entire original ODE by it, we will be able to integrate and solve using 'exact ODE' methods. This is shown as:
$$ \mu \left(\frac{1}{x}\right)dx + \mu \left(1+x^2y^2\right)dy = 0$$
$$ \frac{\partial}{\partial y} \left(\mu\left(\frac{1}{x}\right) \right) = \frac{\partial}{\partial x} \left(\mu \left(1+x^2y^2\right) \right)$$
Now expanding by chain rule, we get:
$$\mu_y \left(\frac{1}{x}\right) = \mu_x \left(1+x^2y^2\right) + \mu \left(2xy^2\right)$$
Now here is where I'm stuck. We want to avoid dealing with a PDE, so we try to stick to good old ODE techniques by assuming that $\mu$ is either a function of only x or only y. 
Let's first assume that $\mu$ is only a function of y. The following will then be true.
$$ \mu_x = 0$$
$$ \mu_y \left(\frac{1}{x} \right) = \mu \left(2xy^2 \right)$$
$$ \frac{d\mu}{\mu} = 2x^2y^2 dy$$
By looking at the right hand side, we see that it just won't work - x and y are related, so we can't have that integral.
Now let's assume that $\mu$ is only a function of x. The following will then be true.
$$ \mu_y = 0$$
$$ \mu_x \left(1+x^2y^2\right) = -\mu \left(2xy^2\right)$$
$$ \frac{d\mu}{\mu} = \frac{-2xy^2}{1+x^2y^2} dx$$
And, once again, if you look at the right hand side, we have an integral that we can't immediately work out, just as in the previous case.
 A: The method of undetermined coefficients can be used to obtain a integrating factor.
We find $\mu$ in form
$$\mu=\frac{1}{\sum_{i,j=0}^2A_{i,j}x^iy^j}$$
We get integrating factor
$$\mu=\frac{1}{2x^2y^2+2x^2y+x^2+2}$$
and exact solution
$$\ln\left(2y^2+2y+1+\frac{2}{x^2}\right)-2y=C$$
A: Assume you're looking for a solution $y(x)$, then mathematic gives:
$$ \frac{-1}{x + x^3 y^2} = \frac{ dy}{dx} \implies  \frac{\exp (-2 y)}{8 x^2} + \frac{ \exp (-2y) }{16} (2 y^2 +2y +1) = Const $$ 
A: It looks to me like we can't find an integrating factor which depends only on $x$ or only on $y$ (in general, $\mu$ will be a function of only one variable just in some special cases, so this is not entirely surprising). 
For any equation of the form $p(x,y)dx + q(x,y)dy = 0$,
in order to be able to find $\mu := \mu(x)$ it must be the case that
\begin{equation}
\frac{\frac{\partial p}{\partial y} - \frac{\partial q}{\partial x}}{q}
\end{equation}
is a function of $x$ only. If it is, we can set
\begin{equation}
\mu(x) = \exp\left(\int\frac{\frac{\partial p}{\partial y} - \frac{\partial q}{\partial x}}{q}dx\right).
\end{equation}
Here I get
\begin{equation}
\frac{\frac{\partial p}{\partial y} - \frac{\partial q}{\partial x}}{q} = 
\frac{-2xy^2}{1 + x^2y^2},
\end{equation}
which clearly depends on $y$. Then we can't have a $\mu$ which depends only on $x$. To have a $\mu$ which depends only on $y$, we must have
\begin{equation}
\frac{\frac{\partial q}{\partial x} - \frac{\partial p}{\partial y}}{p}
\end{equation}
be only a function of $y$. If that's the case, we can set
\begin{equation}
\mu(y) = \exp\left(\int \frac{\frac{\partial q}{\partial x} - \frac{\partial p}{\partial y}}{p} dy\right).
\end{equation}
Here I get
\begin{equation}
\frac{\frac{\partial q}{\partial x} - \frac{\partial p}{\partial y}}{p} = \frac{2xy^2}{\frac{1}{x}} = 2x^2y^2,
\end{equation}
which clearly depends on $x$. Then we can't have $\mu$ depend on $y$ only.
As a result, we must have $\mu$ depend on both $x$ and $y$. 
