Question

求不定积分∫(1-x)/(9-4x^2) dx需要过程～

Answer 1

(1-x)/(9-4x^2)=a/(3+2x)+b/(3-2x)
=[a(3-2x)+b(3+2x)]/(9-4x^2)
所以3a-2ax+3b+2bx=1-x
3a+3b=1
2b-2a=-1
a=5/12,b=-1/12
(1-x)/(9-4x^2)
=(5/12)*1/(3+2x)-(1/12)*1/(3-2x)

所以∫(1-x)/(9-4x^2) dx
=(5/12)∫1/(3+2x)dx-(1/12)∫1/(3-2x)dx
=(5/24)∫1/(3+2x)d(3+2x)+(1/24)∫1/(3-2x)d(3-2x)
=(5/24)*ln|3+2x|+(1/24)*ln|3-2x|+C

Answer 2

把（1-x）放到d里面，变成d
（x-1/2*x^2）,再配平

Answer 3

(1-x)/(9-4x²)
=5/12(2x+3) + 1/12(2x-3)
∫(1-x)/(9-4x²) dx
=∫[5/12(2x+3) + 1/12(2x-3)]dx
=5/24 *ln|2x+3| + 1/24 *ln|2x-3| +C

Answer 4

先化简

(1-x)/(9-4x^2)=5/12(2x+3)+1/12(2x-3)

5/12∫1/2x+3dx + 1/12 ∫1/2x-3dx

这就简单了
剩下的交给你自己了

good luck for you！！

Answer 5

还是不给过程