Question

求不定积分(x^2-a^2)dx/［(x^2+a^2)^2］

Answer 1

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

原积分=∫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