答：
设t=√(x-2)>0,x=t^2+2,dx=2tdt
原式
=∫{x+1/[x√(x-2)]}dx
=∫{t^2+2+1/[(t^2+2)t]} 2tdt
=∫(2t^3+4t)dt+2∫ 1/(t^2+2)dt
=(1/2)t^4+2t^2+(2/√2)arctan(t/√2)+C1
=(1/2)(x-2)^2+2(x-2)+√2arctan[√(x-2)/√2]+C2
=(1/2)(x-2)^2+2x+√2arctan√(x/2-1)+C3
=(x^2)/2+√2arctan√(x/2-1)+C