Answer
$$\frac{{{{\left( {x - 1} \right)}^2}}}{2}\ln x - \frac{{{x^2}}}{4} + x - \frac{1}{2}\ln \left| x \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {x - 1} \right)\ln x} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln x{\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \left( {x - 1} \right)dx{\text{ then }}v = \frac{{{{\left( {x - 1} \right)}^2}}}{2} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {x - 1} \right)\ln x} dx = \left( {\ln x} \right)\frac{{{{\left( {x - 1} \right)}^2}}}{2} - \int {\frac{{{{\left( {x - 1} \right)}^2}}}{2}\left( {\frac{1}{x}dx} \right)} \cr
& {\text{simplifying}} \cr
& \int {\left( {x - 1} \right)\ln x} dx = \frac{{{{\left( {x - 1} \right)}^2}}}{2}\ln x - \int {\frac{{{x^2} - 2x + 1}}{{2x}}dx} \cr
& \int {\left( {x - 1} \right)\ln x} dx = \frac{{{{\left( {x - 1} \right)}^2}}}{2}\ln x - \int {\left( {\frac{x}{2} - 1 + \frac{1}{{2x}}} \right)dx} \cr
& {\text{integrating}} \cr
& \int {\left( {x - 1} \right)\ln x} dx = \frac{{{{\left( {x - 1} \right)}^2}}}{2}\ln x - \int {\left( {\frac{x}{2} - 1 + \frac{1}{{2x}}} \right)dx} \cr
& \int {\left( {x - 1} \right)\ln x} dx = \frac{{{{\left( {x - 1} \right)}^2}}}{2}\ln x - \frac{{{x^2}}}{4} + x - \frac{1}{2}\ln \left| x \right| + C \cr} $$
