求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]

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求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]令x=atanu所以

求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]
求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]

求不定积分(x^2-a^2)dx/[(x^2+a^2)^2]
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