TITLE: Understanding the differential $dx$ when doing $u$-substitution
QUESTION [17 upvotes]: I just finished taking my first year of calculus in college and I passed with an A.  I don't think, however, that I ever really understood the entire $\frac{dy}{dx}$ notation (so I just focused on using $u'$), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important.
Take this problem for an example:
$$ \int 2x (x^2+4)^{100}dx $$
So solving this...
$$u = x^2 + 4 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}$$
And so now I'd have: $\displaystyle \int (u)^{100}du$ which is $\displaystyle \frac{u^{101}}{101} + C$ and then I'd just substitute back in my $u$.
I'm so confused by all of this.  I know how to do it from practice, but I don't understand what is really happening here.  What happens to the $du$ between rewriting the integral and taking the anti-derivative?  Why is writing the $dx$ so important?  How should I be viewing this in when seeing an integral?

REPLY [2 votes]: To understand integration by substitution, you can just use the chain rule in reverse:
\begin{equation}
\int f(g (x)) g'(x) dx  = F (g (x)) + C,
\end{equation}
where $ F $ is an anti derivative of $ f $.  To check this, just take the derivative of the right hand side using the chain rule.