Question

求不定积分∫Inx/(1-x)^2dx

Answer 1

∫Inx/(1-x)^2dx
=∫Inxd（1/(1-x)）
=lnx/(1-x)-∫1/(1-x)d(lnx)
=lnx/(1-x)－∫1/(x(1-x))dx
＝lnx/(1-x)－∫(1/x+1/(1-x))dx
＝lnx/(1-x)－lnx+ln|1－x|+C

Answer 2

∫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