Evaluate $$\int \left(\frac{\ln(x)}{x^2-x\ln(x)}\right)^2(\ln(x)-1)\, dx$$
Now, my attempt was to a add and substract $+1$ and $-1$ like this : $\int (\frac{lnx-1+1}{x^2-xlnx})^2(lnx-1) dx$ . Then, $\int (\frac{lnx-1}{x^2-xlnx}+\frac{1}{x^2-xlnx})^2(lnx-1) dx$ and $2\int \frac{lnx-1}{(x^2-xlnx)^2} dx$ (I've tried to make it look simpler) but I am getting stuck . Any other ideas will be appreciated ! ( wolfram gives an ugly result )
Note
$$\begin{align} &\int \left(\frac{\ln x}{x^2-x\ln x}\right)^2(\ln x-1)\, dx\\ =&\int \left(\frac{\frac{\ln x}x}{1-\frac{\ln x}x}\right)^2\frac{\ln x-1}{x^2}\, dx\\ =&-\int \frac{ \left(\frac{\ln x}x\right)^2 }{\left(1-\frac{\ln x}x\right)^2}d(\frac{\ln x}{x})\\ =&-\int \frac{t^2}{(1-t)^2}dt\\ \end{align}$$ where $t=\frac{\ln x}x$.