求不定积分∫1/x√(x^2-2x+3) dx

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求不定积分∫1/x√(x^2-2x+3)dx求不定积分∫1/x√(x^2-2x+3)dx求不定积分∫1/x√(x^2-2x+3)dx设x-1=√2tanudx=√2(secu)^2du原积分=∫√2(

求不定积分∫1/x√(x^2-2x+3) dx
求不定积分∫1/x√(x^2-2x+3) dx

求不定积分∫1/x√(x^2-2x+3) dx
设x-1=√2tanu
dx=√2(secu)^2du
原积分=∫√2(secu)^2du/[(1+√2tanu)√2secu]
=∫du/(√2sinu+cosu)
=(1/√3)∫du/cos(u-t)
=(1/√3)∫sec(u-t)du
=(1/√3)ln|sec(u-t)+tan(u-t)|+C
=(1/√3) {ln|√3√[(x-1)^2+2] +|x-3||-lnx} +C

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