Reverse Product Rule ODE I don't understand the reverse product rule.
When using integrating factors to solve the ODE below, 
$$\frac{dy}{dx} + \frac{2x}{1+x^2} y = \frac{4}{(1+x^2)^2}$$
We have integrating factor $1+x^2$,
so we multiply by both sides and receive
$$\left(\frac{dy}{dx} + \frac{2x}{1+x^2} y\right) (1+x^2) = (1+x^2)\cdot \frac{4}{(1+x^2)^2}$$
Which, by the reverse product rule is the same as:
$$ \frac{d}{dx} (1+x^2)y = \frac{4}{1+x^2}$$
RHS is straight forward, however I do not understand how we obtain LHS.
Could someone please explain for this case and general case?
 A: One way to think of the product rule is that you are "unpacking" something. Similarly, the reverse product rule would have you "repacking" something.
Just doing algebra, you should find $$\left(\frac{dy}{dx} + \frac{2x}{1+x^2} y\right) (1+x^2)=\frac{dy}{dx}(1+x^2)+2xy$$ Notice the "$2x$" quantity in the equation. Here is the tricky part: We want to rewrite it in a way that makes it more useful to us, like $\frac{d}{dx}(1+x^2)=2x$. This makes our equation look like $$\frac{dy}{dx}(1+x^2)+2xy = \frac{dy}{dx} (1+x^2)+\frac{d}{dx}(1+x^2)y \\ \approx u'v+v'u \\ = (uv)'$$ Ideally you can now see the traces of the product rule in the final result, as we have the derivative of one function ($y$) times a second function ($1+x^2$), plus the derivative of the second function times the first function. $$ \frac{dy}{dx} (1+x^2)+\frac{d}{dx}(1+x^2)y = \frac{d}{dx}[y(1+x^2)]$$ Hopefully that approach makes sense. Again, this process can be tricky because you often have to create or look for something that isn't there. You'll get the hang of it with practice.
A: Suppose you have an equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$, whose integrating factor is $M(x) = e^{\int P(x) dx}$. Note that the derivative of $M(x)$ is $M(x) P(x)$ (check this yourself!). Multiplying the equation through by $M(x)$, you get $M(x) \frac{dy}{dx} + M(x) P(x) y = M(x) Q(x)$, or in other words (and dropping the "(x)" everywhere) $M \frac{dy}{dx} + \frac{dM}{dx} y = MQ$. The left hand side is now just $\frac{d}{dx} (My)$.
A: You're taking too regimented an approach.  Silly recognize that the general first-order equation
$$y' + f(x) y = g(x)$$
is soluble as follows.  Multiply both sides by the factor
$$\exp{\left (\int^x dx' \, f(x') \right )} $$
The derivative of this term is
$$ f(x) \exp{\left (\int^x dx' \, f(x') \right )} $$
So that, upon multiplication, the equation becomes
$$\frac{d}{dx} \left [y \exp{\left (\int^x dx' \, f(x') \right )} \right ] = g(x) \exp{\left (\int^x dx' \, f(x') \right )} $$
which is easily soluble.  In your case, $f(x) = 2 x/(1+x^2)$, so that the equation becomes
$$\frac{d}{dx} \left [y (1+x^2) \right ] = \frac{4}{1+x^2}  \implies y = \frac{C+4 \arctan{x}}{1+x^2}$$
