Question

计算下列积分∫上限是2下限是1(根号下(x^2-1)dx/x)

Answer 1

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

Answer 2

令√(x²-1)=u，则x²-1=u²，xdx=udu，u：0-->√3
∫[1-->2] √(x²-1)/x dx
=∫[1-->2] x√(x²-1)/x² dx
=∫[0-->√3] u*u/(u²+1) du
=∫[0-->√3] (u²+1-1)/(u²+1) du
=∫[0-->√3] 1 du-∫[0-->√3] 1/(u²+1) du
=u-arctanu [0-->√3]
=√3-arctan(√3)
=√3-π/3