The Substitution rule I am attempting to learn about the substitution rule and I can't make sense of what Stewart is trying to say. "To find this integral we use the problem solving strategy of introducing something extra. Here the something extra is a new variable, we change from the variable x to a new variable u. Suppose that we let u be the quantity under the root sing 1, $u=1+x^2$ Then the differential of u is du=2xdx. Notice that if the dx in the notation for an integral were to be interpreted as a differnetial then the differential 2xdx would occur in 1 and so formally without justifying our calculation we could write" The rest doesn't really matter, I just don't understand what is going on at all.
The differential of u is $.5(1+x^2)(2x)$ not what he has.
1: $\int 2x \sqrt{1+x^2}$
 A: The $u$ he is referring to is $1+x^2$, not $\sqrt{1+x^2}$.
I prefer the following explanation for ``$u$-substitution'' (which avoids any $u$'s at all):
The basic method for finding an antiderivative is to recall a corresponding derivative rule:
if you know a basic derivative rule, then you know a basic antiderivative rule.  
Here, we regard a function arising from an application of the chain rule as a ``basic derivative''.
Given one of these, it is easy to find its antiderivative so long as you recognize that it is a derivative resulting from using the chain rule. 
What expressions result from applying the chain rule? Well, they have the form of a product of a composition of functions with the derivative of the inner function of the composition.
For example, for the integral
$$
\int 2x\cos(x^2)\,dx, 
$$
the required antiderivative is $\sin x^2+C$.  Why? Well, you just need to see it; which you would if you've mastered differentiation and the use of the chain rule (my apologies if this sounds condencending, I didn't mean for it to).  
For the integral you give above $\int 2x\sqrt{1+x^2}\,dx$,  you should recognize that
${- 2\over3}(1+x^2)^{3/2}+C$ would work.
With $u$ substitution you're thinking "the integral
$$\int2x\sqrt{1+x^2}\,dx$$ 
is the same as the integral $$\int \sqrt u \,du$$ 
as long as $u$ is replaced by $1+x^2$ after evaluating the latter integral"
This $u$-substitution business is just a way of keeping track of things (which you'll eventually see after practicing with the method), especially when the integrand   is the product of a  composition  of functions with a function that is ''almost'' the derivative of the inner function of the composition.   
I hope this helps. I could elaborate on the last paragraph if you think that would help further.
A: I'll probably get stoned to death for such a clumsy answer, but here's how I do substitution problems:
Basically what you need to do is to get a just a single variable under the square root - that's the $u$ you are substituting. What you want to do is to find a substitution of a form that after derivation produces a differential that will "nullify" the $2x$ from $2x\sqrt{1 + x^2}$
So you go like this:
$$
\begin{align*}
u =& 1 + x^2\\
du =& 2x \ dx\\
dx =&\frac{du}{2x}
\end{align*}
$$
So now when you subsitute:
$$
\begin{align*}
\int \frac{2x}{2x}\sqrt{u}\ du\\
\int 1\sqrt{u}\ du\\
\int \sqrt{u}\ du
\end{align*}
$$
So the integral is:
$$
\int \sqrt{u}\ du= \frac{2}{3}u^{\frac{3}{2}} = \frac{2}{3}(1+x^2)^{\frac{3}{2}}
$$
