C_1'=\frac{\left|
\begin{array}{cc}
 0 & x \\
 \frac{1}{x^2} & 1
\end{array}
\right|}{\left|
\begin{array}{cc}
 -\ln  x & x \\
 -\frac{1}{x} & 1
\end{array}
\right|}=\frac{-\frac{1}{x}}{-\ln  x+1}=\frac{1}{x(\ln  x-1)}\Longrightarrow C_1=\int \frac{1}{x({\ln  x}-1)} \, dx=\ln \left|1-\ln  x \right|+D_1