\(I=\int_{1}^{2}xdx-2\int_{1}^{2}\frac{lnx}{x^2}dx=\frac{x^2}{2}\bigg |_{1}^{2}-2\int_{1}^{2}\frac{lnx}{x^2}dx=\frac{3}{2}-2\int_{1}^{2}\frac{lnx}{x^2}dx\) Tính \(J=\int_{1}^{2}\frac{lnx}{x^2}dx\) Đặt \(u=lnx,dv=\frac{1}{x^2}dx\). Khi đó \(du=\frac{1}{x}dx,v=-\frac{1}{x}\) Do đó \(J=-\frac{1}{x}lnx\bigg |_{1}^{2}+\int_{1}^{2}\frac{1}{x^2}dx\) \(J=-\frac{1}{x}lnx-\frac{1}{x^2}\bigg |_{1}^{2}=-\frac{1}{2}ln2+\frac{1}{2}\) Vậy \(I=\frac{1}{2}+ln2\)
