哪位大神解一下,越详细越好,谢谢了。
1个回答
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x³-1=(x-1)(x²+x+1)
(3x+1)/(x³-1)=(3x+1)/[(x-1)(x²+x+1)]
利用标准拆项公式
(Ax+B)/(x²+x+1) + C/(x-1)
=(Ax²-Ax+Bx-B+Cx²+Cx+C)/[(x-1)(x²+x+1)]
=[(A+C)x²+(-A+B+C)x-B+C]/[(x-1)(x²+x+1)]
=(3x+1)/[(x-1)(x²+x+1)]
得A+C=0
-A+B+C=3
-B+C=1
B+2c=3
求出
C=4/3
B=1/3
A=-4/3
得(-4x+1)/[3(x²+x+1)] + 4/[3(x-1)]
原积分
=∫(-4x+1)/[3(x²+x+1)] + 4/[3(x-1)] dx
=-∫(4x-1)/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-∫(4x+2-3)/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-∫2(2x+1)/[3(x²+x+1)] dx + ∫3/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-2∫d(x²+x+1)/[3(x²+x+1)] + ∫1/[(x+1/2)²+3/4] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (4/3)∫1/[(4/3)(x+1/2)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (4/3)∫1/[(2x/√3+1/√3)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (2√3/3)∫d(2x/√3 +1/3)/[(2x/√3+1/√3)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (2√3/3)arctan(2x/√3 +1/3) + (4/3)ln(x-1) +C
(3x+1)/(x³-1)=(3x+1)/[(x-1)(x²+x+1)]
利用标准拆项公式
(Ax+B)/(x²+x+1) + C/(x-1)
=(Ax²-Ax+Bx-B+Cx²+Cx+C)/[(x-1)(x²+x+1)]
=[(A+C)x²+(-A+B+C)x-B+C]/[(x-1)(x²+x+1)]
=(3x+1)/[(x-1)(x²+x+1)]
得A+C=0
-A+B+C=3
-B+C=1
B+2c=3
求出
C=4/3
B=1/3
A=-4/3
得(-4x+1)/[3(x²+x+1)] + 4/[3(x-1)]
原积分
=∫(-4x+1)/[3(x²+x+1)] + 4/[3(x-1)] dx
=-∫(4x-1)/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-∫(4x+2-3)/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-∫2(2x+1)/[3(x²+x+1)] dx + ∫3/[3(x²+x+1)] dx + ∫4/[3(x-1)] dx
=-2∫d(x²+x+1)/[3(x²+x+1)] + ∫1/[(x+1/2)²+3/4] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (4/3)∫1/[(4/3)(x+1/2)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (4/3)∫1/[(2x/√3+1/√3)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (2√3/3)∫d(2x/√3 +1/3)/[(2x/√3+1/√3)²+1] dx + (4/3)ln(x-1)
=(-2/3)ln(x²+x+1) + (2√3/3)arctan(2x/√3 +1/3) + (4/3)ln(x-1) +C
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