Đặt \(\left\{\begin{array}{l} {u=\ln x} \\ {{\rm d}v=\left(x-2\right){\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=\frac{1}{x} {\rm d}x} \\ {v=\frac{x^{2} }{2} -2x} \end{array}\right. .\)
Khi đó: \(\int \left(x-2\right)\ln x{\rm d}x =\left(\frac{x^{2} }{2} -2x\right)\ln x-\int \left(\frac{x}{2} -2\right) {\rm d}x=\left(\frac{x^{2} }{2} -2x\right)\ln x-\frac{x^{2} }{4} +2x+C.\)