这题比较简单，把分母用因式分解分成简单的式子就可以了，∫1/(x²-x-2)dx＝∫1/(x-2)(x+1)dx＝1/3∫1/(x-2)-1/(x+1)dx＝1/3[ln(x-2)-ln(x+1)]+C＝1/3ln(x-2)/(x+1)+C。
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, => B=-1/3
1/(x^2-x-2)
≡ A/(x-2) + B/(x+1)
≡ (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)|]|(0->3)
=(1/3)[ ln(1/4) - ln2 ]
=-ln2
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024