Question

∫（lnx）/(1-x)^2 d（x） 求不定积分

Answer 1

∫ lnx / (1 - x)² dx，分部积分法
= ∫ lnx d[1 / (1 - x)]
= lnx * 1 / (1 - x) - ∫ 1 / (1 - x) d(lnx)
= lnx / (1 - x) - ∫ 1 / [x(1 - x)] dx ...①
设 1 / [x(1 - x)] = A/x + B/(1 - x)，
由待定系数法得：A = 1、B = 1
= lnx / (1 - x) - ∫ [1 / x + 1 / (1 - x)] dx
= lnx / (1 - x) - ∫ 1 / x dx + ∫ 1 / (1 - x) d(1 - x)
= lnx / (1 - x) - lnx + ln(1 - x) + C
= xlnx / (1 - x) + ln(1 - x) + C

Answer 2

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