有理函数的积分,待定系数法
∫(2x-1)/(x-2)(x-3) dx=∫A/(x-2)+B/(x-3) dx
所以A(x-3)+B(x-2)=2x-1
A+B=2
-3A-2B=-1解得A=-3,B=5
∫(2x-1)/(x-2)(x-3) dx=∫A/(x-2)+B/(x-3) dx
=-3∫dx/(x-2)+5∫dx/(x-3)
=-3ln|x-2|+5ln|x-3|+C