Find the Integrating Factor of $(1-2x^2y^2-4xy^3)dx+(2-2x^3y-4x^2y^2)dy=0$. 
Show that the differential equation 
  $(1-2x^2y^2-4xy^3)dx+(2-2x^3y-4x^2y^2)dy=0$ 
  is not exact, but admits integrating factor $\mu=\mu(xy)$. Find $\mu$ and solve the equation.

With the method I used, I got $\mu(xy)=e^{-x^2y^2}$, but then, I need to integrate the function $e^{-x^2y^2}$.
Any suggestions?
 A: Multiplying through by your integrating factor gives
$\displaystyle e^{-x^2y^2}\big(1-2x^2y^2-4xy^3\big)dx+e^{-x^2y^2}\big(2-2x^3y-4x^2y^2\big)dy=0$, 
and $\displaystyle f_x=e^{-x^2y^2}\big(1-2x^2y^2-4xy^3\big)$ and $\displaystyle f_y=e^{-x^2y^2}\big(2-2x^3y-4x^2y^2\big)$ gives
$\hspace{.3 in}\displaystyle f(x,y)=e^{-x^2y^2}\big(x+2y\big)$.
Therefore $\displaystyle e^{-x^2y^2}\big(x+2y\big)=C$ gives the solution.
A: $$(1-2x^{ 2 }y^{ 2 }-4xy^{ 3 })dx+(2-2x^{ 3 }y-4x^{ 2 }y^{ 2 })dy=0\\ \left( 1-2x{ y }^{ 2 }\left( x+2y \right)  \right) dx+\left( 2-2y{ x }^{ 2 }\left( x+2y \right)  \right) dy=0$$
dividing both side to $ (x+2y)$ we get $$\\ \left( \frac { 1 }{ x+2y } -2x{ y }^{ 2 } \right) dx+\left( \frac { 2 }{ x+2y } -2y{ x }^{ 2 } \right) dy=0\\ \left( \frac { dx }{ x+2y } +\frac { 2dy }{ x+2y }  \right) -\left( 2x{ y }^{ 2 }dx+2y{ x }^{ 2 }dy \right) =0\\ d\left( \ln { \left| x+2y \right|  }  \right) -d\left( { x }^{ 2 }{ { y }^{ 2 } } \right) =0\\ d\left( \ln { \left| x+2y \right|  } -{ x }^{ 2 }{ { y }^{ 2 } } \right) =0\\ \ln { \left| x+2y \right|  } -{ x }^{ 2 }{ { y }^{ 2 } }=C$$
