dx/(x^2+x-2)定积分上限是3下限是2怎么算.

Question

dx/(x^2+x-2) 定积分 上限是3 下限是2 怎么算.

Answer 1

x^2+x-2=(x+2)(x-1),
1/(x^2+x-2)=A/(x+2)+B/(x-1)(A,B为常数)
1/(x^2+x-2)=[A(x-1)+B(x+2)]/(x^2+x-2)=[(A+B)x+2B-A]/(x^2+x-2)
A+B=0,2B-A=1,B=1/3,A=-1/3,
1/(x^2+x-2)=(1/3)[1/(x-1)-1/(x+2)]
∫ dx/(x^2+x-2)=(1/3) ∫ [1/(x-1)-1/(x+2)]dx=(1/3) (ln|x-1|-ln|x+2|)+C=(1/3) ln|(x-1)/(x+2)|)+C,
∫<2,3> dx/(x^2+x-2)=(1/3) ln|(x-1)/(x+2)|)|<2,3>=(1/3)[ln(2/5) -ln(1/4)]=(1/3)ln(8/5)