<s>[INST] 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...)
 [/INST] General way:
$$ x^n \cdot x^{-n} = 1$$
$$ nx^{n-1} x^{-n} + x^n \frac{d}{dx} x^{-n}=0$$
$$ \frac{n}{x^{n+1} } + \frac{d}{dx} x^{-n} = 0$$
Or,
$$ \frac{d}{dx}x^{-n} = \frac{-n}{x^{n+1} }$$
integrating both sides with $x$,
$$ x^{-n} +C = \int \frac{-n}{x^{n+1} } dx$$
Or,
$$ \frac{-n}{x^n} +C'  = \int \frac{1}{x^{n+1} } dx$$
</s>