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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。
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。
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