f(x)=(xlnx)'=1+lnx
∫xf(x)dx=∫x(1+lnx)dx
=∫xdx+∫xlnxdx
=x^2/2+∫lnxd(x^2/2)+C
=x^2/2+lnx*x^2/2-∫x/2 dx+C
=1/4*x^2+1/2*x^2lnx+C