How can I get general solution $y(x)$ in next differential equation : $(1+ xy)dx + (1-xy)dy = 0$? I used mathematica, and got solution, but that solution was too dirty and including lots of hard functions.
Is there any way to get solution for this equation - 
$$(1+ xy)dx + (1-xy)dy = 0$$

Approach $1$:
$(1+xy)~dx+(1-xy)~dy=0$
$(xy-1)\dfrac{dy}{dx}=xy+1$
$\left(y-\dfrac{1}{x}\right)\dfrac{dy}{dx}=y+\dfrac{1}{x}$
Let $u=y-\dfrac{1}{x}$ ,
Then $y=u+\dfrac{1}{x}$
$\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{1}{x^2}$
$\therefore u\left(\dfrac{du}{dx}-\dfrac{1}{x^2}\right)=u+\dfrac{1}{x}+\dfrac{1}{x}$
$u\dfrac{du}{dx}=\left(1+\dfrac{1}{x^2}\right)u+\dfrac{2}{x}$
This belongs to an Abel equation of the second kind.
Approach $2$:
$(1+xy)~dx+(1-xy)~dy=0$
$(xy+1)\dfrac{dx}{dy}=xy-1$
$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=x-\dfrac{1}{y}$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-\dfrac{1}{y}$
$u\dfrac{du}{dy}=\left(1-\dfrac{1}{y^2}\right)u-\dfrac{2}{y}$
This belongs to an Abel equation of the second kind.
