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
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