TITLE: Why treating $\dfrac{\mathrm d y}{\mathrm dx}$ as a fraction gives correct answer?
QUESTION [8 upvotes]: Solve: $$\int x(2-3x)^{11} \, dx $$
The book I am following uses a weird technique to solve this. I  am having trouble understanding why it works. 
Let: 
$$ u = 2-3x$$
$$ x = \frac{2-u}{3} $$
$$ \frac{du}{dx} = -3 $$
Book Technique: 
$$dx = - \frac{du}{3}$$
$$ \int \frac{2-u}{3} u^{11} - \frac{du}{3}$$
$$ -\frac{1}{9} \int 2u^{11} - u^{12} du$$ 
My problem with this is that dy/dx is not a fraction but the limit of one  as  such the terms dy and dx on there own are in a sense meaningless and cannot be manipulated algebraically. For this reason, I solved it with a more formal technique: 
$$ \frac{dy}{du} = (\frac{2-u}{3})(u^{11}) = \frac{2u^{11}}{3} - \frac{u^{12}}{3} $$ 
Given, the reverse chain rule (i.e u-substitution) 
$$ u = g(x),  y = f(u) $$
$$ \int \frac{du}{dx} \frac{dy}{du} dx = \int \frac{dy}{du} du$$
so: 
$$ -\frac{1}{3} \int -3 x(2-3x)^{11} dx = \int \frac{dy}{dx} du   $$
$$ -\frac{1}{9}  \int 2u^{11} - u^{12} du $$
My question is thus: 
The technique used by the book is at most an approximation, i.e by assuming that $dy/dx$ is fraction we are taking an assumption that will not always hold, but in this case it seemingly does. Is this correct way of looking at why the 'book technique' works?

REPLY [8 votes]: So the question is why we have
$$\int f(x) dx = \int f(x(u))\frac{dx}{du}du$$
where $x(u)$ is an invertible function?
Let
$$F(x)=\int f(x)dx$$
and
$$G(u)=\int f(x(u))\frac{dx}{du}du$$
By definition of indefinite integral (anti-derivative), chain rule and inverse function theorem, we have
$$\frac{d}{dx}G(u(x))=\frac{dG}{du}\frac{du}{dx}=f(x(u))\frac{dx}{du}\frac{du}{dx}=f(x)$$
Hence
$$G(u(x))=\int f(x)dx =F(x)$$
In other words we have
$$\int f(x) dx = \int f(x(u))\frac{dx}{du}du$$
and symbolically it is as if the $du$ cancels.
From the above one can see that the origin of the "cancellation" comes from
$$\frac{du}{dx}\frac{dx}{du}=1$$ 
But this is not due to "cancellation" but is by inverse function theorem.