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令√(1+x)=t,x=t^2-1,dx=2tdt,
∫√(1+x)/xdx=2∫t^2/(t^2-1)dt
=2∫(x^2-1+1)/(x^2-1)dx
=2x+∫2/(x^2-1)dx
=2x+∫[1/(x-1)-1/(x+1)]dx
=2x+ln|(x-1)/(x+1)|+C
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