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...)

$$ \frac{1}{x^2} = e^{\ln \left ( \frac{1}{x^2} \right ) }  = e^{\ln \left ( x^{-2} \right ) } = e^{-2\ln (x  ) } = (e^{-2})^{\ln(x)} $$
$$ \text{when} ~~a \in \mathbb{R^+}: ~~~\int a^{\ln(x)} dx = \int \frac1x x^{\ln(a) + 1} dx \\ u=\ln x \Rightarrow du = \frac1x dx \\ = \int e^{(\ln(a) + 1)u} \\v = ( \ln(a) + 1)u \Rightarrow du = \frac{1}{( \ln(a) + 1)} dv \\ = \frac{1}{ \ln(a) + 1} \int e^v dv = \frac{1}{\ln(a) + 1} e^v+C= \frac{1}{\ln(a) + 1}e^{(\ln(a) + 1)u} + C = \frac{e^{(\ln(a) + 1)u}}{\ln(a) + 1} + C = \frac{x^{\ln( a ) + 1}}{\ln(a) + 1} + C $$
Substituting $a = e^{-2}$ we get:
$$ \frac{x^{-2 + 1}}{-2 + 1} + C = - \frac{1}{x} + C$$
