Solving the ODE $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$ Question: 

Solve the ODE given below:
  $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$   


My try: 
The equation is not separable because a function of $x$ is added to a function of $y$.
($y+y\sqrt{x^2y^4+1}$)
Also, it is not linear with respect to $x$ or $y$, because it has the term$\sqrt{x^2y^4+1}$.
On the other hand, it's not a complete ODE because $\frac{d}{dy}(y(1+\sqrt{x^2 y^4+1})) \neq \frac{d}{dx}(2x)$
I also tried homogenous ODE's. But this ODE is not homogenous.  It doesn't seem to be a Clero DE either. 

Any idea?
Thanks in advance.
 A: $$y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$$
$$2xyy'=-y^2(1+\sqrt{x^2 y^4+1})$$
$u(x)=xy^2 \quad\to\quad u'=y^2+2xyy'=y^2-y^2(1+\sqrt{x^2 y^4+1})=-y^2\sqrt{x^2 y^4+1})$
$$xu'=-xy^2\sqrt{x^2 y^4+1})=-u\sqrt{u^2+1}$$
$$\frac{u'}{u\sqrt{u^2+1}}=-\frac{1}{x}$$
$$\int\frac{du}{u\sqrt{u^2+1}}=-\int\frac{dx}{x}$$
$$\ln|u|-\ln|1+\sqrt{u^2+1}|=-\ln|x|+\text{constant}$$
$$\frac{u}{1+\sqrt{u^2+1}}=\frac{c}{x}$$
$$u=\frac{2cx}{x^2-c^2}$$
$$y^2=\frac{u}{x}=\frac{2c}{x^2-c^2}\quad\to\quad y=\pm\sqrt{\frac{2c}{x^2-c^2}}$$
Bringing back into the ODE shows that it agrees.
A: Well, we have:
$$\text{y}\cdot\left\{1+\sqrt{1+x^2\cdot\text{y}^4}\right\}\space\text{d}x+2x\space\text{d}\text{y}=0\tag1$$
Let $\text{y}\left(x\right)=\frac{\text{p}\left(x\right)^\frac{1}{4}}{\sqrt{x}}$, which gives:
$$\frac{x\cdot\text{p}\space'\left(x\right)+2\cdot\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}{2\sqrt{x}\cdot\text{p}\left(x\right)^\frac{3}{4}}=0\tag2$$
So, for $\text{p}\space'\left(x\right)$:
$$\text{p}\space'\left(x\right)=-\frac{2\cdot\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}{x}\tag3$$
Divide both sides by $\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}$:
$$\frac{\text{p}\space'\left(x\right)}{\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}=-\frac{2}{3}\tag4$$
Integrate both sides with respect to $x$:
$$\int\frac{\text{p}\space'\left(x\right)}{\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}\space\text{d}x=\int-\frac{2}{3}\space\text{d}x\tag5$$
Which gives:
$$\ln\left|\frac{1-\sqrt{1+\text{p}\left(x\right)}}{1+\sqrt{1+\text{p}\left(x\right)}}\right|=\text{C}-2\ln\left|x\right|\tag6$$
Taking $\exp$ of both sides, gives:
$$\left|\frac{1-\sqrt{1+\text{p}\left(x\right)}}{1+\sqrt{1+\text{p}\left(x\right)}}\right|=\frac{\text{C}}{\left|x\right|^2}\tag7$$
Set $\text{y}\left(x\right)=\frac{\text{p}\left(x\right)^\frac{1}{4}}{\sqrt{x}}$ back:
$$\left|\frac{1-\sqrt{1+x^2\cdot\text{y}\left(x\right)^4}}{1+\sqrt{1+x^2\cdot\text{y}\left(x\right)^4}}\right|=\frac{\text{C}}{\left|x\right|^2}\tag8$$
A: Setting $$y(x)=\frac{\sqrt{u(x)}}{x}$$ after simplifying we get
$$\frac{du(x)}{dx}=-\frac{\left(\sqrt{u(x)+1}\right)u(x)}{x}$$
and then integrate
$$\int\frac{\frac{du(x)}{dx}}{\left(\sqrt{u(x)+1}-1\right)u(x)}dx=-\int \frac{1}{x}dx$$ and we get
$$\frac{1}{2}\log\left(1-\sqrt{1+u(x)}\right)-\frac{1}{2}\log\left(1+\sqrt{1+u(x)}\right)-\frac{1}{-1+\sqrt{1+u(x)}}=\log(x)+C$$
