第1个回答 2017-12-26
采用换元积分法,倒代换.
设x=1/t,则：dx=d(1/t)=-t^2dt
∫{1/[x^8*(1+x^2)]} dx
= ∫{1/[（1/t）^8*(1+(1/t)^2)]} d(1/t)
= ∫{t^10/（(1+t^2)d(1/t)
=∫(t^8(1+t^2)-t^8)/（(1+t^2)d(1/t)
=∫(t^8(1+t^2)-t^6(1+t^2)+t^4(1+t^2)-t^2(1+t^2)+t^2)/(1+t^2)d(1/t)
=∫[(t^8-t^6+t^4-t^2+t^2/(1+t^2)]d(1/t)
=∫t^8d(1/t)-∫t^6d(1/t)+∫t^4d(1/t)-∫t^2d(1/t)+∫t^2/(1+t^2)d(1/t)
=∫-t^6dt+∫t^4dt-∫t^2d/t+∫dt+∫1/(1+t^2)dt
=-1/7t^7)+1/5t^5-1/3t^3+t+arctant+C
= -1/(7x^7)+1/(5x^5)-1/(3x^3)+1/x+arctan(1/x)+C
设x=1/t,则：dx=d(1/t)=-t^2dt
∫{1/[x^8*(1+x^2)]} dx
= ∫{1/[（1/t）^8*(1+(1/t)^2)]} d(1/t)
= ∫{t^10/（(1+t^2)d(1/t)
=∫(t^8(1+t^2)-t^8)/（(1+t^2)d(1/t)
=∫(t^8(1+t^2)-t^6(1+t^2)+t^4(1+t^2)-t^2(1+t^2)+t^2)/(1+t^2)d(1/t)
=∫[(t^8-t^6+t^4-t^2+t^2/(1+t^2)]d(1/t)
=∫t^8d(1/t)-∫t^6d(1/t)+∫t^4d(1/t)-∫t^2d(1/t)+∫t^2/(1+t^2)d(1/t)
=∫-t^6dt+∫t^4dt-∫t^2d/t+∫dt+∫1/(1+t^2)dt
=-1/7t^7)+1/5t^5-1/3t^3+t+arctant+C
= -1/(7x^7)+1/(5x^5)-1/(3x^3)+1/x+arctan(1/x)+C
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