Question

一道高数积分题求助在线等，

求解这道题做法

Answer 1

简单计算一下即可，答首肢芹历案如者首世图所示

Answer 2

x^2-x-2=(x-2)(x+1)
let
1/(x^2-x-2)≡A/(x-2)+B/(x+1)
=>
1≡A(x+1)+B(x-2)
x=2，=>A=1/3
x=-1，族燃=>B=-1/3
1/(x^2-x-2)
≡A/(x-2)+B/(x+1)
≡信哗(1/3)[1/(x-2)-1/(x+1)]
∫(0->3)dx/(x^2-x-2)
=(1/3)∫(0->3)[1/(x-2)-1/(x+1)]dx
=(1/3)[ln、(x-2)/(x+1)、兆坦虚]、(0->3)
=(1/3)[ln(1/4)-ln2]
=-ln2。