∫x/(x+2)(x+3)dx
=∫ x ×[1/(x+2) -1/(x+3)] dx
=∫ x/(x+2)-x/(x+3) dx
=∫1-2/(x+2)-1+3/(x+3)dx
=∫ 3/(x+3)-2/(x+2) dx
=3ln|x+3|-2ln|x+2|+C请问=∫1-2/(x+2)-1+3/(x+3)dx这一步是什么意思？x/(x+2)
=[(x+2)-2]/(x+2)
=(x+2)/(x+2) -2/(x+2)
=1-2/(x+2)能看懂吗

实际上用待定系数法
直接可把x/(x+2)(x+3)化成3/(x+3)-2/(x+2)