Integral of $1/x^2$ without power rule

Another way is to substitute $x=e^t \implies dx = e^t dt $:

$$\int \frac{1}{e^{2t}} e^t dt = \int e^{-t} dt = -e^{-t} + C = -\frac 1x +C$$

Here, only the fact that $e^x$ is its own derivative is used.

Updated:

I found a new second substitute much faster:

Method $-1$

Let $x=\tan t$, then we have

$$\int \dfrac {1}{x^2} dx=\int \dfrac{1}{\sin ^2(t)} dt=-\cot (t)+C=-\cot (\arctan x)+C=-\dfrac 1 x+C$$

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Method $-2$

How about this substitution?

Let $x=\dfrac {1}{\ln t}$, then we have $$\int \dfrac {1}{x^2} dx=-\int \dfrac{1}{t} dt=-\ln t+C=-\dfrac 1x+C.$$

$$I=\int \frac{dx}{x^2} $$ Let $x=\sin(t)$ to make $$I=\int \cot (t) \csc (t) \,dt=-\csc (t)$$