灵鹊做题网∫(u/(1+u-u^2-u^3)) du,求不定积分
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∫(u/(1+u-u^2-u^3)) du,求不定积分
∫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
再问： 最后三步有点不懂
再答： =(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
再问： 为什么 -(1/2)∫du/(1+u)^2 = +(1/2)*(1/(1+u))
再答： -(1/2)∫du/(1+u)^2 = -(1/2)∫(1+u)^(-2)d(1+u) =(1/2)(1+u)^(-1) =(1/2)(1/(1+u))
再问： 太感谢了、刚开始学微积分有点不熟练、尤其是分式的、终于明白了！谢谢！
=∫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
再问： 最后三步有点不懂
再答： =(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
再问： 为什么 -(1/2)∫du/(1+u)^2 = +(1/2)*(1/(1+u))
再答： -(1/2)∫du/(1+u)^2 = -(1/2)∫(1+u)^(-2)d(1+u) =(1/2)(1+u)^(-1) =(1/2)(1/(1+u))
再问： 太感谢了、刚开始学微积分有点不熟练、尤其是分式的、终于明白了！谢谢！
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