令x=2tanu
dx=2(secu)^2du
原积分=∫[2secu/(8(tanu)^3)] 2(secu)^2du
=∫du/[2(sinu)^3]
=(1/2)∫(cscu)^3du
=(1/2)∫[1+(cotu)^2]cscudu
=(1/2)∫cscudu+(1/2)∫cscu(cotu)^2du
=(1/2)ln|cscu-cotu|-(1/2)∫cscudcscu
=(1/2)ln|cscu-cotu|-(1/4)(cscu)^2+C
=(1/2)ln|√(x^2+4)-2|-(1/2)lnx-(1/4)[(x^2+4)/x^2]+C
=(1/2)ln|√(x^2+4)-2|-(1/2)lnx-(1/x^2)+C
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