Question

求不定积分x/(根号x-1)dx

Answer 1

令根号x-1=t
x=t²＋1
dx=2tdt
原式=∫ (t²＋1)/t · 2tdt
=2∫(t²＋1)dt
=2/3t³＋2t＋c
=2/3 （根号x-1）³＋2(根号x-1)+c

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Answer 2

解：令　√x-1=t 则x=(t+1)² dx=2(t+1)dt
∫x/(√x-1)dx=∫(t+1)²/t*2(t+1)dt
=2∫(t³+3t+3+1/t)dt
=2t³/3+3t² +6t+2lnt+C