用分部积分法求不定积分∫1n(x+1)dx

Question

如题所述

Answer 1

看看对不对

Answer 2

∫ln(x+1) dx
=xln|x+1| - ∫x/(x+1) dx
=xln|x+1| - ∫[ 1- 1/(x+1)] dx
=xln|x+1| - x+ln|x+1| +C本回答被提问者采纳