Apparent contradiction when working out $\int \frac{dx}{x\ln |x|}$

Question

Considering the integral $\int \frac{dx}{x\ln |x|}$ I understand the u-substitution way to proceed by substituting $u=\ln |x|; du=dx/x$ and solving. However, I began by trying integration by parts and came up with the following, $u=\frac{1}{\ln |x|}, u’=\frac{-1}{x(\ln |x|)^2}, v’=1/x, v=\ln |x|$ such that $$\int \frac{dx}{x\ln |x|}=\frac{\ln |x|}{\ln |x|}-\int \frac{-1}{x\ln |x|}dx$$ And, if we factor the -1 from the integral on the RHS and then subtract the integral from both sides, haha, presto $0=1$. Help please, I can’t spot my error. Does integration by parts just fail here, and why doesn’t it work out?