MHB Int ( dx / x ^2 ) = - 1 / x, a new proof.

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The discussion presents a proof for the integral of 1/x², concluding that ∫(dx/x²) = -1/x. The proof utilizes the relationship between integration and differentiation, highlighting that these operations are inverses of each other. Participants note the clever use of notation, particularly the identification of the integral operator with the identity operator. There is also a mention of the limitations of these operators regarding distributive properties with multiplication. Overall, the thread emphasizes the mathematical relationship and rules governing integration and differentiation.
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Proposition. $\displaystyle \int\frac{dx}{x^2}=-\dfrac{1}{x} $

Proof. $$\begin{aligned}\int\frac{dx}{x^2}&=\int\dfrac{dx}{x\cdot x} \\&=\int\dfrac{d\;\not x}{x\cdot \not x}\\&=\left(\int d\right)\frac{1}{x}\\&=id\left(\frac{1}{x}\right) \\&=\frac{1}{x}\end{aligned}$$ Now, using the well-known sign's rule:$\displaystyle \int\frac{dx}{x^2}=-\dfrac{1}{x}\qquad \square $
 
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Fernando Revilla said:
Now, using the well-known sign's rule...

Yes... very well-known (as things usually... are... when someone writes that).
 
I like how $\int \text{d}=\text{id}$.
It's true, since these operators are each others inverses.
And for instance $\int (\text{d}x) = (\int \text{d})x = x \color{silver}{+ C}$
It's just a pity that these operators are not distributive with multiplication. ;)
 
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