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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: $\int 2x \sqrt{1+x^2}$

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    @Jordan Might I suggest you take a look at The Calculus Lifesaver videos - http://press.princeton.edu/video/banner/ - specifically video 12 addresses u-substitution.2011-11-09

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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.

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    I don't underst$a$nd this at all, I think I just have to wait until someone can explain it to me.2011-11-08
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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}} $