Answer
\[\ln|\ln x-1|+C\]
Work Step by Step
Let \[I=\int\frac{dx}{x\ln x-x}\]
\[I=\int\frac{dx}{x(\ln x-1)}\]
Substitute $\;\;\ln x-1=t$ ____(1)
\[\;\;\;\;\;\;\Rightarrow \frac{1}{x}dx=dt\]
\[I=\int\frac{dt}{t}\]
\[I=\ln|t|+C\]
$C$ is constant of integration
Using (1)
\[I=\ln|\ln x-1|+C\]
Hence $\;\;I=\ln|\ln x-1|+C.$
