Chọn phương án D.

\begin{eqnarray*}
&f(x)&=xf'(x)+\ln x\\
\Leftrightarrow&-\ln x&=xf'(x)-f(x)\\
\Leftrightarrow&-\dfrac{\ln x}{x^2}&=\dfrac{xf'(x)-x'f(x)}{x^2}\\
\Leftrightarrow&-\dfrac{\ln x}{x^2}&=\left[\dfrac{f(x)}{x}\right]'\\
\Leftrightarrow&\displaystyle\int\limits_{1}^{\mathrm{e}}\left(-\dfrac{\ln x}{x^2}\right)\mathrm{\,d}x&=\displaystyle\int\limits_{1}^{\mathrm{e}}\left[\dfrac{f(x)}{x}\right]'\mathrm{\,d}x\\
\Leftrightarrow&\displaystyle\int\limits_{1}^{\mathrm{e}}\left(-\dfrac{\ln x}{x^2}\right)\mathrm{\,d}x&=\dfrac{f(x)}{x}\bigg|_1^\mathrm{e}=\dfrac{f(\mathrm{e})}{\mathrm{e}}-f(1)
\end{eqnarray*}

Đặt $\begin{cases}
u=\ln x\\ v'=-\dfrac{1}{x^2}
\end{cases}\Rightarrow\begin{cases}
u'=\dfrac{1}{x}\\ v=\dfrac{1}{x}.
\end{cases}$

Khi đó ta có
\begin{eqnarray*}
&\dfrac{\ln x}{x}\bigg|_1^\mathrm{e}-\displaystyle\int\limits_{1}^{\mathrm{e}}\dfrac{1}{x^2}\mathrm{\,d}x&=\dfrac{f(\mathrm{e})}{\mathrm{e}}-1\\
\Leftrightarrow&\dfrac{1}{\mathrm{e}}+\dfrac{1}{x}\bigg|_{1}^{\mathrm{e}}&=\dfrac{f(\mathrm{e})}{\mathrm{e}}-1\\
\Leftrightarrow&\dfrac{1}{\mathrm{e}}+\dfrac{1}{\mathrm{e}}-1&=\dfrac{f(\mathrm{e})}{\mathrm{e}}-1\\
\Leftrightarrow&\dfrac{2}{\mathrm{e}}&=\dfrac{f(\mathrm{e})}{\mathrm{e}}\\
\Leftrightarrow&2&=f(\mathrm{e}).
\end{eqnarray*}