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

[Fernando Revilla]

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 $

 

[Ackbach]

Now, using the well-known sign's rule...

Yes... very well-known (as things usually... are... when someone writes that).

 

[I like Serena]

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. ;)