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求不定积分:∫(x+1)dx/[x√(x-2)]
解: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
解: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
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