Integration Grandmaster - P4

The substitution $u = xe^{-\frac1x}$ gives $\frac{du}{dx} \; = \; \frac{x+1}{x}e^{-\frac1x}$ and hence $u\frac{du}{dx} \; = \; (x+1)e^{-\frac2x}$ and so the desired integral becomes $A \; = \; \int_0^{2/\sqrt{e}} ue^u\,du \; = \; \Big[(u-1)e^u\Big]_0^{2/\sqrt{e}} \; = \; 1 + \left(\frac{2}{\sqrt{e}} - 1\right)e^{2/\sqrt{e}}$ Thus $\lfloor 10^4A \rfloor = \boxed{17166}$ .