Using integration by parts, would the following be correct?

$\displaystyle \int(2x-1)\ln x\,dx$

$\displaystyle u = \ln x \Rightarrow du = \frac{1}{x}\,dx$

$\displaystyle dv = (2x - 1)\,dx \Rightarrow v = \int 2x - 1\,dx = x^2 - x$

$\displaystyle \int(2x - 1)\ln x\,dx = \ln x(x^2 - x) - \int{\frac{x^2 - x}{x}}\,dx$

$\displaystyle = \ln x(x^2 - x) - \int{x - 1}\,dx$

$\displaystyle = \ln x(x^2 - x) - \frac{x^2}{2} + x + C$