定积分问题 0到1∫ln(1+x)dx等于多少？求过程，谢谢

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原式=∫(0,1)ln(x+1)d(x+1)
=(x+1)ln(x+1) (0,1)-∫山唯(0,1)(x+1)dln(x+1)
=(x+1)ln(x+1) (0,1)-∫察毁(0,1)(x+1)*1/ln(x+1) dx
=(x+1)ln(x+1) (0,1)-∫败唯备(0,1) dx
=[(x+1)ln(x+1)-x] (0,1)
=2ln2-1

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