令√(x-1)=t,那么x=t^2+1
所以原式=∫1/[(t^2+1)×t]d(t^2+1)
=∫2t/[(t^2+1)×t]dt
=∫2/(t^2+1)dt
=2×arctant +C
=2arctan[√(x-1)] +C
方法就是换元.
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