Chapter 7.2, Problem 64E

Log integrals Use integration by parts to show that for m ≠ −1,

${\displaystyle \int x^m\ln xdx=\frac{x^{m+1}}{m+1}\left(\ln x-\frac{1}{m+1}\right)+C}$

and for m = −1,

${\displaystyle \int \frac{\ln x}{x}dx=\frac{1}{2}{\ln }^{2}x+C.}$