<s>[INST] Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method 
By integrating factor method - 
I put it into the form of $ \frac{dy}{dx} + \frac{4x}{x^2 + 1} \cdot y = \frac{x}{x^2 + 1} $
My integrating factor is $I = (x^2 + 1)^2 $ 
By formula, $ y (x^2 +1)^2 = \int{ (x^3 + x)} dx $
And thus my general solution is $ y = \frac{x^4 + 2x^2 + C}{4(x^2 + 1)^2} $
By separable method is where I got problems with... 
I put it into the form of $\int{ (\frac{x}{x^2 + 1} )} dx = \int{ (\frac{1}{(1-4y)}} dy $
Getting $\frac{1}{2} \ln (x^2 + 1) + C = \frac{1}{4} \ln (1-4y) $
$2 \ln (x^2 +1) + C = \ln (1-4y) $ 
$1-4y = e^{\ln (x^2 + 1)^2} + e^{C} = (x^2 +1)^2 + C $
$ y = \frac{ (x^2+1)^2 + C - 1}{4} $ 
Why is my general solution from separable method very different from the integrating factor method ? I suspect that I have went wrong in the separable method but cannot identify it ... 
 [/INST] Yes, you have.
In separable form ,
$$\displaystyle \int \dfrac{1}{1-4y}~dy=\int\dfrac{x}{x^{2}+1}~dx$$
$\dfrac{-1}{4}\log(1-4y)=\dfrac{1}{2}\log(x^{2}+1)$ You have missed minus sign.
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