Question

求一个不定积分，谢谢了啊~

Answer 1

∫ε(x-ε)^3dε/[(x-ε)^2+z^2]^2
= -∫(x-ε)^4dε/[(x-ε)^2+z^2}^2 +x∫(x-ε)^3dε/[(x-ε)^2+z^2}^2
=-∫[(x-ε)^2-z^2]dε/[(x-ε)^2+z^2]+z^4∫dε/[(x-ε)^2+z^2]^2 -(x/4)∫d(x-ε)^4/[(x-ε)^2+z^2]^2
=-ε-2z^2∫dε/[(x-ε)^2+z^2] -z∫d[(x-ε)/z]/[(x-ε)^2/z^2+1]^2-(x/4)∫d[(x-ε)/z]^4/[(x-ε)^2/z^2+1]^2
=-ε+2zarctan[(x-ε)/z] -(z/2)[arctan[(x-ε)/z] +[(x-ε)z/((x-ε)^2+z^2)]} -(x/4)[2ln|(x-ε)^2/z^2+1|
+2z^2/[(x-ε)^2+z^2] ] +C

x-ε/z=tanu ∫d[(x-ε)/z]/[(x-ε)^2/z^2+1]^2=∫cosu^2du=(1/2)u+(1/2)sinucosu
=(1/2)arctan[(x-ε)/z]+(1/2)[(x-ε)z]/[(x-ε)^2+z^2]

∫d(x-ε)^4/z^4/[(x-ε)^2/z^2+1]^2=∫4tanu^3secu^2du/secu^4
=∫4tanu^3cosu^2du=4∫sinu^3du/cosu
=4∫-(1-cosu^2)dcosu/cosu
=-4∫dcosu/cosu+2(cosu)^2
=4ln|secu|+2(cosu)^2
=2ln|(x-ε)^2/z^2+1|+2z^2/[(x-ε)^2+z^2]