Help with simplifying Differential Equation Original Problem: Solve Differential Equation 
$$\left(x^2 \sec{\left(y+1\right)} \tan (y+1)-3xe^{y+1}+\frac{2yx^{2}}{y^2+1}\right) \frac{dy}{dx}+3x \sec(y+1)-6e^{y+1}+3x \ln{\left(y^2+1\right)}=0$$
I am working on a differential equation and got to this pint when doing the integrating factor: 
$$\frac { x \sec(y+1) \tan(y+1)-3e^{y+1}+\frac{4xy}{y^2+1} } { x^2 \sec(y+1) \tan(y+1)-3x e^{y+1}+\frac{2yx^{2}}{y^2+1} }$$
Can I go any further. I know that if I multiply the top by x all the variables are the same on top and bottom. Then I divide long division and get 2. So is this the answer to this integrating factor?
 A: So we have $Pdx+Qdy=0$, where
$$
P=3x \sec(y+1)-6e^{y+1}+3x \ln{\left(y^2+1\right)}
$$
$$
Q=x^2 \sec{\left(y+1\right)} \tan (y+1)-3xe^{y+1}+\frac{2yx^{2}}{y^2+1}
$$
It is not an exact equation, but may be we can obtain an exact equation multiplying by a suitable factor:
$$
P_y=3x \tan(y+1) \sec(y+1)-6e^{y+1}+3x \frac{2y}{y^2+1}
$$
$$
Q_x=2x \tan(y+1) \sec(y+1)-3e^{y+1}+\frac{4yx}{y^2+1}
$$
Note that
$$
P_y-Q_x=x\tan(y+1)\sec(y+1)-3e^{y+1}+ \frac{2yx}{y^2+1}=\frac{Q}{x}
$$
Then
$$
\frac{\alpha'(x)}{\alpha(x)}=\frac{P_y-Q_x}{Q}=\frac{1}{x}
$$
$$
\frac{d\alpha}{\alpha}=\frac{dx}x
$$
Therefore the factor $\alpha=x$ becomes the equation in an exact equation. It remains to integrate the new equation...
We need to find the potential $g$ of the vector field $(x\,P(x,y),x\,Q(x,y))$:
$$
g(x,y)=\int 3x^2 \sec(y+1)-6x\,e^{y+1}+3x^2 \ln{\left(y^2+1\right)} \,dx +c(y) = \\ 
=x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c(y) = \\ =x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c
$$
Finally, the solutions are given by the equation:
$$
x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c =0
$$
