To integrate by parts, we have to pick a #u# and #dv# such that
#int u dv= uv-int v du#
We can pick what #u# and #dv# are, though we should typically try to pick #u# such that #du# is "simpler". (By the way, #du# just means the derivative of #u#, and #dv# just means derivative of #v#)
So, from #int u dv#, let's pick #u=x# and #dv=e^x#. It is helpful to fill in all of the parts you will use throughout the problem before you start integrating:
#u=x#
#(du)/dx=1# so #du=1dx=dx#
#dv=e^x#
#v=int dv= int e^x dx =e^x#
Now let's plug everything back into the original formula.
#int u dv= uv-int v du#
#int xe^x dx= xe^x-int e^xdx#
You can integrate the last part quite simply now, to get:
#int xe^x dx= xe^x-int e^xdx= xe^x-e^x +c#