求不定积分：∫(x+1)dx/[x√(x-2)]令√(x-2)=u,则x=u²+2,dx=2udu,代入原式得：原式=2∫(u²+3)du/(u²+2)=2∫[1+1/(u²+2)]du=2[u+∫du/(u²+2)]=2{u+∫du/2[(u/√2)²+1]}=2u+(√2)∫d(u/...