Chọn phương án D.

Đặt $\begin{cases}
u=\ln x\\ v'=x
\end{cases}\Rightarrow\begin{cases}
u'=\dfrac{1}{x}\\ v=\dfrac{x^2}{2}
\end{cases}$. Ta có $$\begin{aligned}
\displaystyle\int x\ln x\mathrm{\,d}x&=\dfrac{x^2\ln x}{2}-\int\dfrac{1}{x}\cdot\dfrac{x^2}{2}\mathrm{\,d}x\\
&=\dfrac{x^2\ln x}{2}-\int\dfrac{x}{2}\mathrm{\,d}x\\
&=\dfrac{x^2\ln x}{2}-\dfrac{x^2}{4}+C.
\end{aligned}$$