"Determine the area between the graph of $\displaystyle y=xe^{-x^{2}}$ and the x axis over the interval [0,1]."

my work:

Area = $\displaystyle \int_0^1 \! xe^{-x^{2}} \, dx$

using the integral by parts rule " $\displaystyle \int \! udv = uv - \int \! vdu$ "

u = x | v = $\displaystyle \frac{e^{-x^{2}}}{-2x}$

du = 1 | dv= $\displaystyle e^{-x^{2}}$

thus..

$\displaystyle

\int_0^1 \! xe^{-x^{2}} \, dx = x(\frac{e^{-x^{2}}}{-2x}) - \int \! \frac{e^{-x^{2}}}{-2x} (1)$

This is where I'm like, am I even on the right track?

and if so, how do I integrate $\displaystyle \int \! \frac{e^{-x^{2}}}{-2x}$?

Thanks for the help.