I can't figure out this simplification in a differential equation I was watching PatrickJMT's video on first-order differential equations and while I think I should see what he's doing on the left side here from line one to line two, I just can't. I ran it past my roommate as well without any luck :(
Elaboration: I understand that he's multiplying everything by the integrating factor ($1+x^2$), but how does he compresses it to simply ($\frac{d}{dx}((1+x^2)y$)?

 A: The second line is just the power rule $(uv)' = u'v + uv'$, but performed in reverse. If we expand it out starting from the second line you'll be able to see what he did:
$$
\begin{align}
\frac{d}{dx}\left[(1+x^2)y\right] & = y\frac{d}{dx}(1+x^2) + (1+x^2)\frac{dy}{dx} \\
& = 2xy + (1+x^2)\frac{dy}{dx} \\
& = \left(\frac{dy}{dx} + \frac{2x}{1+x^2}y\right)(1+x^2)
\end{align}
$$
or if you prefer moving from the first line to the second,
$$
\left(\frac{dy}{dx} + \frac{2x}{1+x^2}y\right)(1+x^2) = \frac{4}{(1+x^2)^2}(1+x^2) \\
\frac{dy}{dx}(1+x^2) + 2xy = \frac{4}{1+x^2} \\
y'(1+x^2) + y(1+x^2)' = \frac{4}{1+x^2} \\
\left[y(1+x^2)\right]' = \frac{4}{1+x^2} \\
\frac{d}{dx}\left[(1+x^2)y\right] = \frac{4}{1+x^2}
$$
A: This is done in LHS (using the rule $(uv)' = u'v + uv'$):
$\frac{d}{dx}((1+x^2)y)=(\frac{dy}{dx})(1+x^2)+y\frac{d}{dx}(1+x^2)=(\frac{dy}{dx})(1+x^2)+y(2x)=(1+x^2)\frac{dy}{dx}+y\frac{2x}{1+x^2}(1+x^2)=(1+x^2)(\frac{dy}{dx}+y\frac{2x}{1+x^2})$
A: Use $\frac{d}{dx}[1+x^2]=2x$ on the following:
$(\frac{dy}{dx}+\frac{2x}{x^2+1}y)(1+x^2)=(1+x^2)\frac{dy}{dx}+2xy$
and apply rule for differential of product.
