$I=\int\limits_{1}^{e}e^{x+lnx}dx=\int\limits_{1}^{e}e^x.xdx$Đặt $u=x, dv=e^xdx,$ ta có:
$u'=1$, chọn $v=e^x.$
Khi đó,
$I=x.e^x\bigg|_1^e-\int\limits_{1}^{e}e^xdx=e^{e+1}-e-e^x\bigg|_1^e=e^{e+1}-e-(e^e-e)=e^e(e-1).$
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