Question

∫(u/(1+u-u^2-u^3)) du， 求不定积分

Answer 1

∫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

追问

最后三步有点不懂

追答

=(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

追问

为什么
-(1/2)∫du/(1+u)^2
= +(1/2)*(1/(1+u))

追答

-(1/2)∫du/(1+u)^2
= -(1/2)∫(1+u)^(-2)d(1+u)
=(1/2)(1+u)^(-1)
=(1/2)(1/(1+u))

追问

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！