Question about finding $du$ from $dx$ when integrating by substitution

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Example 1:

$$\int\frac{\tan^2(\ln x)}{x}dx$$

Method 1:

$$\text{Let}\ u=\ln x$$

$$\implies du=d(\ln x)$$

$$\implies du=\ln'(x)dx$$

$$\implies du=\frac{1}{x}dx$$

Method 2:

$$\text{Let}\ u=\ln x\tag{Start}$$

$$\implies \frac{du}{dx}=\frac{1}{x}$$

$$\implies xdu=dx$$

$$\implies du=\frac{dx}{x}\tag{End}$$

Example 2

$$\int\frac{dx}{x\sqrt{x^4-1}}$$

Method 1:

$$\text{Let}\ x^4=\frac{1}{u^2}$$

$$\implies u=\frac{1}{x^2}$$

$$\implies du=d(\frac{1}{x^2})$$

$$\implies du=d(x^{-2})$$

$$\implies du=-2x^{-3}dx$$

Method 2:

$$\text{Let}\ x^4=\frac{1}{u^2}\tag{Start}$$

$$\implies 4x^3=-2u^{-3}\frac{du}{dx}$$

$$\implies 2x^3=-u^{-3}\frac{du}{dx}$$

$$\implies 2x^3dx=-u^{-3}du$$

$$\implies du=-\frac{2x^3dx}{u^{-3}}$$

$$\implies du=-\frac{2x^3dx}{(\frac{1}{x^2})^{-3}}$$

$$\implies du=-2x^{-3}dx\tag{End}$$

Questions:

  1. In both examples 1 and 2, in method 1, I just used this the rule $d(f(t)) = f'(t)dt$. However, in method 2, I used differentiation in both sides, cross multiplication, dividing both sides by $x$, other general rules we use in equations etc. I'm certain that method 1 is correct. In both method 1 and 2, I got the same end expression, so the result of method 2 is also correct. However, are the intermediate lines between the start and the end correct in method 2? I'm asking this because I used cross multiplication and other techniques; I managed to arrive at the correct answer, but I'm anxious whether the intermediate lines are correct to write or not i.e. is LHS=RHS true in those lines.

Related: 1,2

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The reason for why you want to find a $\mathrm{d}u$ from a $\mathrm{d}x$ is in reality because you have the substitution theorem which tells you that, simply put,

$$\int f(u(x))\frac{\mathrm{d}u(x)}{\mathrm{d}x}\mathrm{d}x=\int f(u)\mathrm{d}u.$$

Thus what you do in reality when you do a substitution, i.e. there is no actual fraction which you can cancel things in. However this theorem tells you that you can essentially treat it like a fraction when you do a substitution, but formally it is your second method which is more along the lines of what is actually going on.