Answer
\[x\ln x-x+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int \ln x\:dx\]
\[I=\int \ln x.(1)dx\]
Using integration by parts
\[I=\ln x\int dx-\int \left((\ln x)'\int dx\right)dx\]
\[I=x\ln x-\int\left(\frac{1}{x}\right)xdx\]
\[I=x\ln x-\int dx\]
\[I=x\ln x-x+C\]
Where $C$ is constant of integration
Hence ,
\[I=x\ln x-x+C\]
