Question

这两个不定积分怎么求

第六第五题

Answer 1

望有所帮助

Answer 2

变换分式的题目

Answer 3

6.x/(x²-2x-3)=(x-3+3)/(x-3)(x+1)=1/(x+1)+3/(x-3)(x+1)=1/(x+1)+3/4*[1/(x-3)-1/(x+1)]=3/4*1/(x-3)+1/4*1/(x+1)所以原式=3/4*ln|x-3|+1/4*ln|x+1|+C
前面加负号

追答

Answer 4

(6)
∫xdx/(3+2x-x^2)
= -(1/2) ∫ (2-2x)/(3+2x-x^2) dx + ∫dx/(3+2x-x^2)
= -(1/2)ln|3+2x-x^2| + ∫dx/(3+2x-x^2)
= -(1/2)ln|3+2x-x^2| - ∫dx/(x^2-2x-3)
= -(1/2)ln|3+2x-x^2| - ∫dx/[(x-3)(x+1)]
= -(1/2)ln|3+2x-x^2| -(1/4) ∫[1/(x-3)-1/(x+1)] dx
= -(1/2)ln|3+2x-x^2| -(1/4)ln|(x-3)/(x+1)| + C
(5)
∫(x-2)dx/((3+2x-x^2)
=-(1/2)∫(2-2x)/(3+2x-x^2) dx-∫dx/(3+2x-x^2)
=-(1/2)ln|3+2x-x^2| + ∫dx/(x^2-2x-3)
=-(1/2)ln|3+2x-x^2| + ∫dx/[(x-3)(x+1)]
=-(1/2)ln|3+2x-x^2| + (1/4) ∫[1/(x-3)-1/(x+1)] dx
=-(1/2)ln|3+2x-x^2| + (1/4) ln|(x-3)/(x+1)| +C