1/（x+2）(x+3)=1/(x+2)-1/(x+3)
所以原式=∫1/(x+2)d(x+2)-∫1/(x+3)d(x+3)
=ln|x+2|-ln|x+3|+C
=ln|(x+2)/(x+3)|0到无穷
x=0,ln|(x+2)/(x+3)|=ln(2/3)
x→+∞,(x+2)/(x+3)=(1+2/x)/(1+3/x),极限=1,所以ln|(x+2)/(x+3)|极限=0
所以原式=0-ln(2/3)=ln(3/2)