Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of luck
\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}
This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write
\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}
you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.
The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that
\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}
and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

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.
