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求不定积分dx/[(x-1)^4根号(x^2-2x)]
人气:345 ℃ 时间:2020-01-27 22:57:34
解答
√(x^2-2x)=t,x^2-2x=t^2 (x-1)^2=t^2+1,(x-1)dx=tdt,代入得:
∫dx/[(x-1)^4根号(x^2-2x)]
=∫tdt/t(t^2+1)^(5/2)
=∫dt/(t^2+1)^(5/2)
=t(2t^2+3)/3(t^2+1)^(3/2)+C (查积分表)
最后代t
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