Stanford Math tournament(2014) Question

can you make it more better

x=f(x)e^f(x).

now take f(x)=y.

x=ye^y.

∫ydx=G(x).

i write ∫ydx as ∫y.1dx now by using integration by parts.

iget ∫y.1dx as y.∫1dx-∫[(dy/dx).∫1dx].dx.

now by cancelling dx

we get G(x)=yx-∫xdy.

we know that x=ye^y so ∫xdy=∫(y.e^y)dy now again by using integration by parts we get ∫y.e^ydy as( y.e^y-e^y).

so we see that G(x)=xy-(y.e^y-e^y).

no you can find out that y(e)=1,y(0)=0,hence G(e)=e and G(0)=1.

so the answer to the question i posted is e-1. Sorry Nahom assefs the solution i posted now is much clear hope it helps you.