I'm learning more about how to solve integrals with substitution. Before, I was relying on the formula

$\int f(g(x))g'(x)dx = \int f(u)du$ where $u=g(x)$

But I've noticed another way that it is done, where $g'(x)$ is unnecessary. It seems to involve looking at the problem from another perspective. Is this correct?

$\int (2x+3)^3xdx = \int (2x+3)^3 \cdot \frac{(2x+3)-3}{2}dx$

I can now integrate with respect to $(2x+3)$ instead of $x$. If $u=(2x+3)$, then

\begin{align*} \int (2x+3)^3 \cdot \frac{(2x+3)-3}{2}dx&=\int (u)^3\frac{u-3}{2}du\\ &=\int\frac{u^4}{2}-\frac{3u^3}{2}du\\ &=\frac{1}{10}u^5-\frac{3}{8}u^4+c\\ &=\frac{1}{10}(2x+3)^5-\frac{3}{8}(2x+3)^4+c \end{align*}

Is that correct?

I see that the solution would be to see that

$g(x)=u=2x+3$

$g'(x)=\frac{du}{dx}=2$

$dx=\frac{du}{2}$

I can follow this. However, the last of the three lines there bothers me. I would like to avoid manipulating derivatives like quotients. It's something that has caused me problems before. I avoid using something when I don't understand how it really works. Does that make sense?

EDIT:

I'm thinking that

$\int f(g(x))dx=\lim_{n \to \infty} \sum_{i=1}^n f((g^{-1}(x))_i^*) \cdot \frac{g^{-1}(b)-g^{-1}(a)}{n}$

..I'm thinking out loud so that might not be right, but that's the kind of approach I want to take to solve this.

-
Basically it's correct, but you miss a factor of $1/2$ when you change from $x$ to $u$, because $dx=\frac{1}{2}du$. –  Paul Mar 1 '12 at 5:42
I don't know what $g^{-1}(x)^*_i$ means, and, anyway, what if $g$ is not invertible? And do you have anything to say about the two answers that have been posted? –  Gerry Myerson Mar 1 '12 at 22:59

You can always check an answer to an integration problem by differentiating. If you differentiate your result, you will get $(2x+3)^4-3(2x+3)^3$. (If you get something else, you forgot to use the Chain Rule.)

Take out the common factor of $(2x+3)^3$. We get $(2x+3)^3((2x+3)-3)$, which is $(2x+3)^3(2x)$, not what you were trying to integrate.

Essentially, you were doing a substitution: in fact you wrote $u=2x+3$ explicitly. You just didn't bother with the $du=2\,dx$, or equivalently $dx=\frac{du}{2}$. That accounts for the fact that your answer is off by a factor of $2$.

Remark: Do remember that one can always check by differentiating! When one does a complicated integration, slippage (a missing minus sign, a wrong constant) is all too frequent, at least for me. Differentiating will almost always detect an error.

-

You replaced $dx$ with $du$, but $dx\ne du$, given that $u=2x+3$.

EDIT: Going up to your first paragraph, the $g'(x)$ is important, because it reminds you that you are really doing a chain rule. $(f(g(x)))'=f'(g(x))$ isn't correct; you have to have $(f(g(x)))'=f'(g(x))g'(x)$. And it's exactly that $g'(x)$ that you miss when you replace $dx$ with $du$. Don't think of it as treating a derivative like a fraction; think of it as a bookkeeping device to help you keep track of the $g'(x)$ that has to be there because of the chain rule.

-