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- Jan 29, 2012

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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 $