Found the above integral in the old posts and figured I would play around with it using double substitution and integration of partial derivatives. So here goes.
$$\int \frac {dx}{x^2-2x}$$
$$= \int\frac {dx}{x(x-2)}$$
Then let
${u=x,du=dx}$
and
${v=x-2,dv=dx}$
Now here is where I stumble on to something or show a lack of competence so you be the judge. If you judge it the later please tell me what I am doing wrong.
$$={\int\int u^{-1}v^{-1}\delta u \delta v}$$ $$=\frac {ln|u|}{v}\frac{ln|v|}{u}$$ substituting back I get $$= \frac {ln|x|}{x-2} \frac {ln|x-2|}{x}+c$$
Did I get this right?? or do I need to go back to the drawing board?
note: I know there are other ways to do this but please critique this approach. Thank you
${EDIT}$
The above math was incorrect. Because ${dy=udv+vdu}$ I left the post unedited to so others may follow the conversation. The math should read:
$${\frac {dx}{f(x)g(x)}}$$
let
${u=f(x), \delta u=g(x)\delta x=v\delta x, then:\delta u/v= \delta x}$
${v=g(x),\delta v=f(x)\delta x=u\delta x, then:\delta v/u=\delta x}$
This only makes sense as partials so
${dx=\delta x_1+\delta x_2}$
then
${dx=\delta u/v+\delta v/u}$
That makes the expression $${\frac{1}{uv}(\delta u/v+\delta v/u) }$$ Distributing $${\frac {1}{uv^2} \delta u + \frac {1}{vu^2} \delta v } $$
then $${\int \frac {dx}{f(x)g(x)}= \int \frac{1}{uv} = \int\frac {1}{uv^2} \delta u + \int \frac {1}{vu^2} \delta v = \frac {lnu}{v^2}+ \frac {lnv}{u^2}+c}$$
Then the original expression $${\int \frac{1}{x(x-2)}={\frac {ln|x|}{(x-2)^2 } + \frac {ln|x-2|}{x^2}+c}}$$
Went back over the algorthim and it looks right. So I am pretty sure this is the answer.
${Edit}$
Never mind this method just won't work.
Why not try using partial fraction decomposition?
$$\int \frac{dx}{x(x-2)} = \int \Big(\dfrac A{x} + \frac B{x-2}\Big) \,dx$$
$$A(x-2) + Bx = 1 \implies -2A = 1, \quad A = - B$$
$$A = -\frac 12, B = \frac 12$$
$$\begin{align} \int \frac{dx}{x(x-2)} & = \frac 12 \int \Big(\dfrac{1}{x - 2} - \frac 1x\Big) \,dx \\ \\ & = \dfrac 12 \left( \ln|x - 2| - \ln |x|\right) + C \\ \\ & = \frac 12 \ln\left|\frac{x-2}{x}\right|+C\end{align}$$