先裂项：
1/[(x^2+1)*(x^2+4)] =1/3*[1/(x²+1)-1/(x²+4)]
∴:∫(0-->+∞）1/[(x^2+1)*(x^2+4)] dx
=∫(0-->+∞）1/3*[1/(x²+1)-1/(x²+4)]dx
=1/3*∫(0-->+∞)1/(x²+1)dx-1/6*∫(0-->+∞)1/[(x/2)²+1) d(x/2)
=1/3 arctanx|(0-->+∞)-1/6arctan(x/2)|(0-->+∞)
=1/3(π/2)-1/6*(π/2)
=π/12