∫上0下e-1 ln(x+1)dx
问题描述:
∫上0下e-1 ln(x+1)dx
答
原式=∫(0~(e-1))ln(x+1)d(x+1)
=(x+1)ln(x+1)(0~(e-1))-∫(0~(e-1))(x+1)dln(x+1)
=(x+1)ln(x+1)(0~(e-1))-x(0~(e-1))
=-e+e-1
=-1