计算下列积分∫上限是2下限是1(根号下(x^2-1)dx/x)

令x = secz,dx = secztanz dzx > 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