令x=atanu
所以tanu=x/a
sinu=x/√x^2+a^2
cosu=a/√x^2+a^2
sin2u= 2ax/x^2+a^2
dx=a(secu)^2du
原積分=∫a^2[(tanu)^2-1]a(secu)^2du /[a^4(secu)^4]
=(1/a)∫[(sinu)^2-(cosu)^2]du
=-(1/a)∫cos2udu
= -sin2u/2a+C
= -x/(x^2+a^2)+C