Question on intuition behind notation in u substitution I have a question concerning u substitution for integrating. Suppose I needed to set $u = 3x^2 + 1$. Then $du = 6x dx$. So why are we incorporating the $dx$ with it? 
I understand that $\displaystyle\frac{du}{dx} = 6x$, but I was under the impression that $\displaystyle\frac{du}{dx}$ could not be thought of as a fraction. Its just a way of symbolizing the derivative of $u$ with respect to $x$. But it seems like it is being treated like a fraction because both sides are being multiplied by $dx$.   
 A: Think of $dx$ as the width of a very, very small movement in the x direction.  Think of $du$ as the corresponding change in u direction (when $u=3x^2+1$).  $du/dx$ is just this ratio.  When $dx$ is an infinitesimally thin slice, then $du/dx$ happens to be the derivative.
So if it helps, think of du and dx as small but positive numbers.  I'm not getting into the weeds of the technicalities here (a field called analysis or real analysis deals with this), but that's the idea for now.
A: You are right to be worried about treating "infinitessimals" as if they were ordinary quatities, to be multiplied using the ordinary rules.  In fact, the subject of hyper-reals was developed to put that sort of manipulation on a rigorous footing, but it turns out that subtlteies emerge.
Instead, you can think of the statement that 
$$
du = g(x) dx
$$
as being shorthand for the statement that for any differentiable $f(x)$, 
$$
\frac{df(x)}{dx} = g(x) \frac{df(x)}{du}
$$
For an expression like $u = 3x^2+1$ the differentiation chain rule says precisely that for any ifferentiable $f(x)$, 
$$
\frac{df(x)}{dx} = 6x \frac{df(x)}{du}
$$
or in our shorthand, that
$$
du = g(x) dx
$$
This is important in doing integrals, because the prpoerty allows you to substitute $u$ for the expression in terms of $x$, and tells you immediately what function of $x$ you need to multiply by to cancel out the effect of the chain rule.  
