Chọn D
\(I=\int _{1}^{e}x\ln x{\rm d}x =\frac{e^{a} +1}{b} \)
Đặt \(\left\{\begin{array}{l} {u=\ln x} \\ {{\rm d}v=x{\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=\frac{1}{x} {\rm d}x} \\ {v=\frac{1}{2} x^{2} } \end{array}\right. \)
Suy ra: \(I=\left. \frac{1}{2} x^{2} \ln x\right|_{1}^{e} -\frac{1}{2} \int _{1}^{e}x{\rm d}x =\left. \frac{1}{2} x^{2} \ln x\right|_{1}^{e} -\left. \frac{1}{4} x^{2} \right|_{1}^{e}\)
\(=\frac{e^{2} }{2} -\frac{e^{2} }{4} +\frac{1}{4} =\frac{e^{2} +1}{4}
\Rightarrow \left\{\begin{array}{l} {a=2} \\ {b=4} \end{array}\right. \Rightarrow \frac{a}{b} =\frac{1}{2} . \)