设 t = √(x-1) 则 x = t²+1,dx = 2tdt
∫1/[1+√(x-1）]dx
=∫1/[1+ t] 2tdt
= 2∫(t+1-1)/(1+ t)dt
= 2t -2∫1/(1+ t)dt
= 2t - 2ln(1+t) + C
= 2√(x-1) - 2ln(1+√(x-1))+ C
请教一高数不定积分题1/（1+根号x-1）dx 的不定积分
设 t = √(x-1) 则 x = t²+1,dx = 2tdt
∫1/[1+√(x-1）]dx
=∫1/[1+ t] 2tdt
= 2∫(t+1-1)/(1+ t)dt
= 2t -2∫1/(1+ t)dt
= 2t - 2ln(1+t) + C
= 2√(x-1) - 2ln(1+√(x-1))+ C