We use integration by parts. Set $u = x$ and $dv/dx = e^x$. This leads to $du/dx = 1$, and $v = e^x$. Then:\begin{align*}\int xe^x \mathop{\mathrm{d}x} &= xe^x - \int e^x \mathop{\mathrm{d}x}\&= xe^x - e^x.\end{align*}
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