$\frac{d}{dx}\left(\frac{\frac{1}{x^2-2x+1}}{\ln\left(x\right)}\right)=\frac{-\left(x^2-2x+1+\left(2x-2\right)x\ln\left(x\right)\right)}{x\left(x^2-2x+1\right)^2\ln\left(x\right)^2}$
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