C_2'=\frac{\left|
\begin{array}{cc}
 -\ln  x & 0 \\
 -\frac{1}{x} & \frac{1}{x^2}
\end{array}
\right|}{\left|
\begin{array}{cc}
 -\ln  x & x \\
 -\frac{1}{x} & 1
\end{array}
\right|}=\frac{-\frac{ \ln x}{x^2}}{-\ln  x+1}=\frac{\ln  x}{x^2(\ln  x-1)}\Longrightarrow C_2=\int \frac{\ln  x}{x^2(\ln  x-1)} \, dx=-\frac{1}{x}+\frac{1}{e}\mathrm{Ei}(1-\ln  x)+D_2