`intlogxdx=?`

Explanation

Integrating by parts, taking log x as the first function and 1 as the second function, we get
`intlogxdx=int(logx*1)dx`
`=(logx)*1dx-int{(d)/(dx)(logx)*int1dx}dx`
`=(logx)*x-int((1)/(x)*x)dx=xlogx-intdx`
`=xlogx-x+C=x(logx-1)+C`.