计算反常积分∫1/(x+2)(x+3)dx 上限是+∞ 下限是0
问题描述：
计算反常积分∫1/(x+2)(x+3)dx 上限是+∞ 下限是0
答
原式=∫[1/(x+2)-1/(x+3)]dx(0≤x+∞)
=[ln(x+2)-ln(x+3)](0≤x+∞)
=ln[(x+2)/(x+3)](0≤x+∞)
=lim(x→+∞)ln[(x+2)/(x+3)]-ln(2/3)
=lim(x→+∞)ln[(1+2/x)/(1+3/x)]-ln(2/3)
=0-ln(2/3)
=ln(3/2)原式=∫[1/(x+2)-1/(x+3)]dx(0≤x+∞)？？为什么？1/(x+2)(x+3)=1/(x+2)-1/(x+3)