Calculus u-substitution: Choosing between $\frac{dx}{du}$ and $\frac{du}{dx}$

Question

I searched and couldn't find any answers to this one. It's probably a dumb question but one that has been troubling me.

Let's say I have an integration: $$\int \frac{1}{3x+2}$$

It seems the correct way to solve it is by choosing $u=3x+2$. So

$$\begin{align}u &= 3x + 2 \\ \frac{du}{dx} &= 3 \\ dx &= \frac{1}{3}du \end{align}$$

and so on the answer comes down to $\frac{1}{3}\ln(3x+2)$

My question is, why do I need to choose it specifically this way: $ \frac{du}{dx} = 3$? If I choose $ \frac{dx}{du} = 3$, I get $dx=3du$ , but it gives me an answer different from the above.

How do I choose which one I use as a denominator and nominator for my u substitution?

Thank you.

Answer 1

We want to simplify the denominator and set $u=u(x)=3x+2$. Here we consider $u=u(x)$ being a function in $x$. \begin{align*} u&=3x+2\\ \color{blue}{\frac{du}{dx}}=\frac{d}{dx}u(x)&=\frac{d}{dx}(3x+2)\color{blue}{=3} \end{align*} On the other hand if we consider $x=x(u)$ as function in $u$ we have \begin{align*} x&=\frac{1}{3}(u-2)\\ \color{blue}{\frac{dx}{du}}=\frac{d}{du}x(u)&=\frac{d}{du}\left(\frac{1}{3}u-\frac{2}{3}\right)\color{blue}{=\frac{1}{3}} \end{align*} So, it is not reasonable to set $\frac{dx}{du}=3$.

Note: It's advisable to write $\int\frac{\color{blue}{dx}}{3x+2}$ which besides some other aspects indicates that $x$ is the integration variable.

Answer 2

You are incorrect in saying: $\frac{dx}{du} = 3$

You don't "choose" this.

This is a direct result of what your u-sub is.

If we let $u = 3x+2$, then differentiating both sides, we get:

$du=3 dx$

If you wanted t o have $\frac{dx}{du}$, following the equation, this must equal $1/3$, not $3$.