Chọn phương án A.

Đặt \(\begin{cases}
u=x\\ v'=\mathrm{e}^x
\end{cases}\Rightarrow\begin{cases}
u'=1\\ v=\mathrm{e}^x
\end{cases}\)

Khi đó $$\begin{aligned}I&=\displaystyle\int\limits_{1}^{2}\left(x\mathrm{e}^x-1\right)\mathrm{\,d}x\\ &=x\mathrm{e}^x\bigg|_1^2-\displaystyle\int\limits_{1}^{2}\mathrm{e}^x\mathrm{\,d}x-\displaystyle\int\limits_{1}^{2}\mathrm{\,d}x\\ &=2\mathrm{e}^2-\mathrm{e}-\left(\mathrm{e}^x+x\right)\bigg|_1^2=\mathrm{e}^2-1.\end{aligned}$$