Understanding U-substitution method to deal with compound integral functions

Question

I am starting to learn Integrals by using u-substitution method and I came across an example in which there is a part I don't quite understand.

Let's suppose I have got this Integral here: $$ \int (3x²+ 2x){e^{(x^3 + x^2)}}dx $$ Then, I set u to be equal to: $$ u = x^3 + x^2 $$

Then the derivative of u is simply:

$$\frac{\partial u}{\partial x}= 3x^2 + x $$

So far, so good, the problem for me is now:

$$(\partial x)\frac{\partial u}{\partial x}= (3x^2 + x)(\partial x) $$

It seems like both sides of the equation were multiplied by $$(\partial x) $$

Therefore becoming this: $$\partial u = (3x^2 + x)(\partial x) $$

It doesn't make any sense whatsoever in my mind: DX is not a number, it cannot be treated as so and be multiplied in both sides to obtain DU. Correct if I am wrong but, DX is the length of that little rectangle that we are summing when tackling integrals by using Riemmans' sum, right?

Could you guys be so kind as to explain to me why we can do this maneuver?

Thanks in advance, I really appreciate it

Answer 1

You can just think of being $$\int f'(x)*e^{f(x)} dx$$ Some people like to call $f'(x) dx$ $df$.

Answer 2

You're not really multiplying both sides by $dx$. The definition of "differential" is

$$dy = f'(x) \; dx.$$

Another symbol for $f'(x)$ is $\frac{dy}{dx},$ so the definition can be written

$$dy = \frac{dy}{dx} \; dx,$$

where it's understood that $dy/dx$ is a symbol for the derivative and not $dy$ divided by $dx$.

So you start with

$$\frac{du}{dx} = 3x^3+2x$$

then multiply both sides by $dx$

$$\frac{du}{dx}\; dx = (3x^3+2x) \; dx.$$

Now the left side is $du$ by definition of differential. You didn't cancel the $dx$'s.