1/(x^3+9)如何积分急

Question

1/(x^3+9) 如何积分 急

Answer 1

∫dx/(x^3+a^3)
=∫dx/[(x+a)(x^2-ax+a^2)]
=∫[(x+a)^2-(x^2-ax+a^2)]dx/[3ax(x+a)(x^2-ax+a^2)]
=∫(x+a)dx/[3ax(x^2-ax+a^2)]-∫dx/[3ax(x+a)]
=(1/3a)∫dx/(x^2-ax+a^2)+(1/3)∫dx/[x(x^2-ax+a^2)]-(1/3)ln[|x|/|x+a|]
=(1/(3a))∫dx/[(x-a/2)^2+3a^2/4] +(1/3)∫dx/[x(x^2-ax+a^2)]-(1/3)ln[|x|/|x+a|]
=(1/(3a))arctan(2x/√3a-1/√3) -(1/3)ln|x|/|x+a|+(1/3)∫dx/[x(x^2-ax+a^2)]
=(1/3a)arctan(2x/√3a-1/√3)-(1/3)ln|x|/|x+a| +(1/6)ln|x^2-ax+a^2|-(a/6)arctan(2x/√3a-1/√3)-(1/3a^2)ln|x|+C
∫dx/(x^3+9)=(1/3^(5/3))arctan(2x/3^(7/6)-1/√3)-(1/3)ln|x|/|x+3^(2/3)|+(1/6)ln|x^2-3^(2/3)x+3^(4/3)|
-[3^(-1/3)/2]arctan(2x/3^(7/6)-1/√3)-(1/3^(7/3))ln|x| +C
∫dx/[x(x^2-ax+a^2)]
=∫xdx/[x^2(x^2-ax+a^2)]=(1/2)∫d(x^2)/[x^2(x^2-ax+a^2)]
=(1/2)∫[x^2-(x^2-ax+a^2)]dx^2/[(ax-a^2)x^2(x^2-ax+a^2)]
=(1/2)∫dx^2/[(ax-a^2)(x^2-ax+a^2)]-(1/2)∫dx^2/[(ax-a^2)x^2]
=(1/a^2)∫axdx/[(a-x)(x^2-ax+a^2)- (1/a)∫dx/[x(a-x)]
=(1/a^2)∫[(x^2-ax+a^2)-(a-x)^2]dx/[(a-x)(x^2-ax+a^2)] -(1/a^2)ln|x|/|x-a|
=(1/a^2)∫dx/(a-x)-∫(a-x)dx/(x^2-ax+a^2)-(1/a^2)ln|x|/|x-a|
=(-1/a^2)ln|x-a)+∫(x-a)dx/(x^2-ax+a^2)-(1/a^2)ln|x|/|x-a|
=(1/2)∫(2x-a)dx/(x^2-ax+a^2)-(1/2)∫adx/(x^2-ax+a^2)-(1/a^2)ln|x|
=(1/2)ln|x^2-ax+a^2|-(a/2)arctan(2x/√3a-1/√3) -(1/a^2)ln|x|