求不定积分∫Inx/(1-x)^2dx

Question

需要过程，谢谢啦

Answer 1

∫Inx/(1-x)^2dx
=-∫Inx/(1-x)^2d(1-x)
=∫Inxd1/(1-x)
=lnx/(1-x)-∫x/(1-x)dx
=lnx/(1-x)+∫(1-x+1)/(1-x)dx
=lnx/(1-x)+∫[1+1/(1-x)]dx
=lnx/(1-x)+x-ln(1-x)+C

Answer 2

用分部积分即可
原式=∫lnxd1/(1-x)
=lnx/(1-x)+∫1/x(x-1)dx
=lnx/(1-x)+ln[x/(x-1)]+c