Chọn D
Đặt \(\left\{\begin{array}{l} {u=\ln x} \\ {dv=xdx} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {du=\frac{1}{x} } \\ {v=\frac{1}{2} x^{2} } \end{array}\right.\)
Ta có: \(\int _{1}^{2}x\ln xdx= \left. \left(\frac{1}{2} x^{2} \ln x\right)\right|_{1}^{2} -\int _{1}^{2}\frac{1}{2} x^{2} .\frac{1}{x} dx=2\ln 2-\frac{1}{2} \int _{1}^{2}xdx=2\ln 2-\left. \left(\frac{1}{4} x^{2} \right)\right| _{1}^{2} =2\ln 2-\frac{3}{4} .\)
Theo bài ra ta có: \(2\ln 2+a=2\ln 2-\frac{3}{4} \Rightarrow a=-\frac{3}{4} .\)