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...)
 A: 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.
A: $$I=\int \frac{dx}{x^2} $$ Let $x=\sin(t)$ to make
$$I=\int \cot (t) \csc (t) \,dt=-\csc (t)$$
A: Let $f(x)=-1/x\;(x>0)$. Then
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac1h\left(\frac1x-\frac1{x+h}\right)=\lim_{h\to0}\frac1{x(x+h)}=\frac1{x^2}.$$Hence, for some constant $c$,
$$\int\frac1{x^2}\,\mathrm dx= -\frac1x+c.$$
A: I thought this answer might be useful to you as it involves simple manipulations.
$$\begin{align} \int \dfrac {1}{x^2}dx=\int \dfrac {1}{x} \times \dfrac {1}{x}dx=\int x^{-1}d(\ln x)=\int e^{\ln x^{-1}}d(\ln x)=\int e^{- \ln x}d(\ln x)=- \int e^{- \ln x}d(- \ln x)=-e^{- \ln x}+C=-\dfrac {1}{x}+C.\end{align}$$
A: You could use $u=x^{-2}.\,v=x$ so $\int x^{-2}dx=x^{-1}+2\int x^{-2}dx\implies\int x^{-2}dx=-x^{-1}+C$.
A: 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.
A: 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
$$\int \dfrac {1}{x^2} dx=-\int \dfrac{1}{t} dt=-\ln t+C=-\dfrac 1x+C.$$
A: $$ \frac{1}{x^2} = e^{\ln \left ( \frac{1}{x^2} \right ) }  = e^{\ln \left ( x^{-2} \right ) } = e^{-2\ln (x  ) } = (e^{-2})^{\ln(x)} $$
$$ \text{when} ~~a \in \mathbb{R^+}: ~~~\int a^{\ln(x)} dx = \int \frac1x x^{\ln(a) + 1} dx \\ u=\ln x \Rightarrow du = \frac1x dx \\ = \int e^{(\ln(a) + 1)u} \\v = ( \ln(a) + 1)u \Rightarrow du = \frac{1}{( \ln(a) + 1)} dv \\ = \frac{1}{ \ln(a) + 1} \int e^v dv = \frac{1}{\ln(a) + 1} e^v+C= \frac{1}{\ln(a) + 1}e^{(\ln(a) + 1)u} + C = \frac{e^{(\ln(a) + 1)u}}{\ln(a) + 1} + C = \frac{x^{\ln( a ) + 1}}{\ln(a) + 1} + C $$
Substituting $a = e^{-2}$ we get:
$$ \frac{x^{-2 + 1}}{-2 + 1} + C = - \frac{1}{x} + C$$
A: $$I=\int \frac 1{x^2} dx$$
Let
$v=x\implies dv=dx$
Let $u.v'=1\implies u=1 \implies u'=0$
$$I=\int\frac{uv'}{v^2}=\int\frac{u'}{v}-\frac{u}{v}=0-\frac{1}{x}=-\frac{1}{x}$$
A: $$I=\int \frac{1}{x^2} \, dx$$
Let $x=\cot(y)$, with $\frac{dx}{dy}=-\csc^2(y)$
$$I=-\int \frac{\csc^2(y)}{\cot^2(y)} \, dy=-\int \sec ^2(y) \, dy=-\tan(y)+C=-\frac{1}{x}+C$$
or
Let $x=\sec(y)$, with $\frac{dx}{dy}=\sec(y)\tan(y)$
$$I=\int \frac{\sec(y)\tan(y)}{\sec^2(y)} \, dy=\int \sin(y) \, dy=-\cos(y)+C=-\frac{1}{x}+C$$
A: 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,
$$ \frac{-n}{x^n} +C'  = \int \frac{1}{x^{n+1} } dx$$
