原积分=∫(1,2)1/x^2×d(x^2-1)^(1/2),令(x^2-1)^(1/2)=u，则原积分=∫(0,√3)du/(u^2+1)
=arctanu(u=√3)-arctanu(u=0)
=π/3.
∫1/[x(x^2-1)^(1/2)]dx
=arccos(1/x)+C
∫(1,2)dx/[x(x^2-1)^(1/2)]
=arccos(1/2)-arccos(1)
令x=sect
原式=∫(0,π/3)dt=π/3
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