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