第1个回答 2012-03-19
令x = secz,dx = secztanz dz
x > 1,z ∈[0,π/2),√(x² - 1) = √(sec²z - 1) = tanz
∫(1~2) √(x² - 1)/x dx
= ∫(0~π/3) tanz/secz * secztanz dz
= ∫(0~π/3) tan²z dz
= ∫(0~π/3) (sec²z - 1) dz
= [tanz - z] |(0~π/3)
= [tan(π/3) - π/3] - [tan(0) - 0]
= √3 - π/3本回答被提问者采纳