求不定积分：分子是1+x,分母是x*（x-2）^0.5

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/√2)/[(u/√2)²+1)]=2u+(√2)arctan(u/√2)+C
=2√(x-2)+(√2)arctan√[(x-2)/2]+C