求不定积分∫(2x^2-5x+5)dx⼀(x-2)(1-x)^2

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2024-11-18 23:23:33
推荐回答(1个)
回答1:

∫(2x^2-5x+5)/[(x-2).(x-1)^2] dx
let
(2x^2-5x+5)/[(x-2).(x-1)^2]≡ A/(x-2) + B/(x-1) + C/(x-1)^2
=>
2x^2-5x+5≡ A(x-1)^2 + B(x-1)(x-2) + C(x-2)
x=1, =>C=-2
x=2, =>A=3
coef. of x^2
A+B=2
3+B=2
B=-1
(2x^2-5x+5)/[(x-2).(x-1)^2]≡ 3/(x-2) - 1/(x-1) - 2/(x-1)^2
∫(2x^2-5x+5)/[(x-2).(x-1)^2] dx
=∫ [3/(x-2) - 1/(x-1) - 2/(x-1)^2] dx
=3ln|x-2| -ln|x-1| +2/(x-1) + C