$\displaystyle \frac{1}{2}xe^{\frac{-x}{2}} $
i used parts and wondering wether im correct.thats to the power of -x/2 (its a bit small)!
ok first of all substutute $\displaystyle u=\frac{x}{2}$
therefore $\displaystyle x=2u$ and $\displaystyle du=\frac{1}{2}dx$
so we have
$\displaystyle 2 \int u e^{-u} du$
using parts once we have
$\displaystyle 2(-ue^{-u}+\int e^{-u} du)$
$\displaystyle 2(-ue^{-u}-e^{-u})+c$
$\displaystyle -2e^{-u}(u+1)+c$
$\displaystyle -2e^{-\frac{x}{2}}(\frac{x}{2}+1)+c$
$\displaystyle -2e^{-\frac{x}{2}}(\frac{x+2}{2})+c$
$\displaystyle -e^{-\frac{x}{2}}(x+2)+c$