∫(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|] +...最后三步有点不懂=(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为什么 -(1/2)∫du/(1+u)^2= +(1/2)*(1/(1+u))-(1/2)∫du/(1+u)^2= -(1/2)∫(1+u)^(-2)d(1+u)=(1/2)(1+u)^(-1)=(1/2)(1/(1+u))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！