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 $$\int \dfrac {1}{x^2} dx=-\int \dfrac{1}{t} dt=-\ln t+C=-\dfrac 1x+C.$$

Let $x=\cos \theta +i\sin \theta$. By De Moivre's Theorem, $$\frac{1}{x^2}=x^{-2}=\cos (-2\theta)+i\sin (-2\theta)=\cos (2\theta)-i\sin(2\theta)$$ Also $$dx=(-\sin \theta+i\cos \theta) d\theta$$ The integral becomes $$\int( \cos 2\theta-i\sin 2\theta)(-\sin \theta+i\cos\theta)d\theta\\ =\int (-\sin \theta\cos 2\theta+i\cos \theta\cos 2\theta+i\sin\theta\sin 2\theta+\cos \theta\sin2\theta)d\theta\\ =\int -\frac{1}{2}(\sin 3\theta-\sin \theta)+i\frac{1}{2}(\cos 3\theta+\cos \theta)+i\frac{1}{2}(\cos\theta-\cos 3\theta)+\frac{1}{2}(\sin 3\theta+\sin\theta)d\theta \\ =\int \sin \theta +i\cos \theta d\theta=-\cos \theta +i\sin\theta +C\\ =-(\cos \theta-i\sin \theta)+C=-(\cos (-\theta)+i\sin (-\theta))+C=-\frac{1}{x}+C$$

You can find $\int x^a\:dx,a\in \mathbb Z$ in a similar manner.