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
令a=√(x-1)
x=a²+1
dx=2ada
原式=∫a/(a²+1)*2ada
=2∫a²/(a²+1)da
=2∫(a²+1-1)/(a²+1)da
=2∫[1-1/(a²+1)]da
=2a-2arctana+C
=2√(x-1)-2arctan√(x-1)+C
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