∫(u/(1+u-u^2-u^3)) du, 求不定积分
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∫udu/[(1+u)-(u^2+u^3)]
=∫udu/[(1+u)^2(1-u)]
=(1/2)∫[(1+u)-(1-u)]du/[(1+u)^2(1-u)]
=(1/2)∫du/[(1+u)(1-u)] -(1/2)∫du/(1+u)^2
=(1/4)∫[(1+u)+(1-u)]du/[(1+u)(1-u)] +(1/2)*(1/(1+u))
=(1/4)ln[|1+u|/|1-u|] +(1/2)*(1/(1+u))+C
=∫udu/[(1+u)^2(1-u)]
=(1/2)∫[(1+u)-(1-u)]du/[(1+u)^2(1-u)]
=(1/2)∫du/[(1+u)(1-u)] -(1/2)∫du/(1+u)^2
=(1/4)∫[(1+u)+(1-u)]du/[(1+u)(1-u)] +(1/2)*(1/(1+u))
=(1/4)ln[|1+u|/|1-u|] +(1/2)*(1/(1+u))+C
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=(1/2)∫[(1+u)-(1-u)]du/[(1+u)^2(1-u)]
=(1/2)∫(1+u)du/[(1+u)^2(1-u)] -(1/2)∫(1-u)du/[(1+u)^2(1-u)]
=(1/2)∫du/[(1+u)(1-u)] -(1/2)∫du/(1+u)^2
=(1/4)∫[(1+u)+(1-u)]du/[(1+u)(1-u)] +(1/2)*(1/(1+u))
=(1/4)∫(1+u)du/[(1+u)(1-u)]+(1/4)∫(1-u)]du/[(1+u)(1-u)]+(1/2)*(1/(1+u)
=(1/4)(-ln|1-u|)+(1/4)(ln|1+u|) +(1/2)*(1/(1+u)
=(1/4)ln[|1+u|/|1-u|] +(1/2)*(1/(1+u))+C
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