∫ 1/((2+x^2)^2)dx
令x=√2tanu,则2+x^2=2(secu)^2,dx=√2(secu)^2du,x：0-->+无穷,u：0-->π/2
原式=∫ 1/(secu)^4*√2(secu)^2du
=√2∫ (cosu)^2du
=√2/2∫ (1+cos2u)du
=√2/2 (u+1/2sin2u) u：0-->π/2
=√2/2*π/2
=√2π/4