求不定积分：∫(x+3)/(x^2-5x+6)dx=

∫(x+3)/(x^2-5x+6)dx
=∫(x+3)/[(x-2)(x-3)]dx
=∫(x-3+6)/[(x-2)(x-3)]dx
=∫{1/(x-2)+6*[(x-2)-(x-3)]/[(x-2)(x-3)]}dx
=∫[1/(x-2)+6/(x-3)-6/[(x-2)]dx
=∫[6/(x-3)-5/[(x-2)]dx
=6ln|x-3|-5ln|x-2|+C
或者直接用待定系数法：
设
(x+3)/[(x-2)(x-3)]=A/(x-2)+B/(x-3)
通分比较对应项系数求A、B即可.