This one gave me some trouble:
2y(x^2-y+x)dx + (x^2-2y)dy=0
I believe the integrating factor is e^(2x) but I may be wrong.
Thanks on advance, sorry about lack of latex, I'll redo it when I get home and am not on my phone.
I was bored so here is more than just the verification of the integrating factor.
$\displaystyle {\partial M\over\partial y}=2x^{2}-4y+2x$
$\displaystyle {\partial N\over\partial x}=2x$
$\displaystyle \frac{{M}_{y}-{N}_{X}}{N(x,y)}=\frac{2x^{2}-4y+2x-2x}{x^{2}-2y}=2$
$\displaystyle {e}^{2\int dx}={e}^{2x}$
$\displaystyle (2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x})dx+(x^{2}{e}^{2x}-2y{e}^{2x})dy$
$\displaystyle f(x,y)=\int (x^{2}{e}^{2x}-2y{e}^{2x})dy=x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)$
$\displaystyle x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)$
$\displaystyle {\frac{d}{dx}}[x^{2}{e}^{2x}y-y^{2}{e}^{2x}+g(x)]=2yx^{2}{e}^{2x}-2y^{2}{e}^{2x}+2yx{e}^{2x}+g'(x)$
$\displaystyle g'(x)=0$
$\displaystyle g(x)=C$
$\displaystyle x^{2}{e}^{2x}y-y^{2}{e}^{2x}+C$