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令√(x-1)=t,则x=t²+1
∫[√(x-1)/x]dx
=∫[t/(1+t²)]d(t²+1)
=∫[2t²/(1+t²)]dt
=2∫[1 -1/(1+t²)]dt
=2t -2arctant +C
=2√(x-1)-2arctan[√(x-1)] +C
∫[√(x-1)/x]dx
=∫[t/(1+t²)]d(t²+1)
=∫[2t²/(1+t²)]dt
=2∫[1 -1/(1+t²)]dt
=2t -2arctant +C
=2√(x-1)-2arctan[√(x-1)] +C
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