令t = x - 1,dt = dx当x = 1,t = 0当x = 2,t = 1原式= ∫(0→1) (t + 1)/√t dt= ∫(0→1) (t/√t + 1/√t) dt= ∫(0→1) (√t + 1/√t) dt= [(2/3)t^(3/2) + 2√t] | (0→1)= (2/3) + (2)= 8/3定积分计算 ∫2(上)1(下)根号x-1/xdx再帮个忙吧令t = √(x - 1)，t² = x - 1，2t dt = dx当x = 1，t = 0当x = 2，t = 1原式= ∫(0→1) t/(1 + t²) • 2t dt= 2∫(0→1) t²/(1 + t²) dt= 2∫(0→1) (t² + 1 - 1)/(1 + t²) dt= 2∫(0→1) [1 - 1/(1 + t²)] dt= 2[t - arctan(t)]| (0→1)= 2{[1 - arctan(1)] - [0 - arctan(0)]= 2[1 - π/4]= 2 - π/2