设√x+1=t
x=t²-1
∫√x+1/xdx=∫t/(t²-1)d(t²-1)
=2∫t²/(t²-1)dt
=2∫(t²-1+1)/(t²-1)dt
=2[∫1dt+∫1/(t²-1)dt]
=2t+∫(t+1-t+1)/(t+1)(t-1)dt
=2t+ln(t-1)-ln(t+1)
设√x+1=t
x=t²-1
∫√x+1/xdx=∫t/(t²-1)d(t²-1)
=2∫t²/(t²-1)dt
=2∫(t²-1+1)/(t²-1)dt
=2[∫1dt+∫1/(t²-1)dt]
=2t+∫(t+1-t+1)/(t+1)(t-1)dt
=2t+ln(t-1)-ln(t+1)