求定积分∫（上2下1）根号下（x^2-1）⼀xdx

解:由题意可得:先求∫√(x^2-1)/xdx的不定积分
令√(x^2-1)=t,又上下限均大于0
所以x=√(t^2+1),dx=t/√(t^2+1)dt
所以∫√(x^2-1)/xdx=∫t/√(t^2+1)*[t/√(t^2+1)]dt
=∫t^2/(t^2+1)dt=∫dt-∫1/(t^2+1)dt
=t-arctant+C将t=√(x^2-1)代人可得
∫√(x^2-1)/xdx=√(x^2-1)-arctan√(x^2-1)+C
然后分别把积分的上下限代人相减可得
∫（上2下1)√（x^2-1）/xdx=√3-π/3

解:设x=sect,
则cost=1/x,dx=sect*tantdt,
且当x=1时，t=0.当x=2时，t=π/3
∴原式=∫(0，π/3)(tant/sect)sect*tantdt
=∫(0，π/3)tan²tdt
=∫(0，π/3)(sec²t-1)dt
=∫(0，π/3)sec²tdt-∫(0，π/3)dt
=∫(0，π/3)d(tant)-∫(0，π/3)dt
=(tant-t)|(0，π/3)
=√3-π/3