简单有理函数积分x^2+5x+6/(x-1)(x^2+2x+3)dx
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发布时间：2023-06-12 00:32
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热心网友
时间：2024-11-13 20:27
待定系数法
。
令(x²+5x+6)/[(x-1)(x²+2x+3)]=a/(x-1)+
(bx+c)/(x²+2x+3)
a(x²+2x+3)+(bx+c)(x-1)=x²+5x+6
(a+b-1)x²+(2a-b+c-5)x+3a-c-6=0
a+b-1=0
2a-b+c-5=0
3a-c-6=0
解得a=2，b=-1，c=0
∫(x²+5x+6)dx/[(x-1)(x²+2x+3)]
=∫[2/(x-1)-x/(x²+2x+3)]dx
=∫[2/(x-1)-½(2x+2)/(x²+2x+3)
+1/(x²+2x+3)]dx
=∫[2/(x-1)]d(x-1)
-½∫[1/(x²+2x+3)]d(x²+2x+3)+(1/√2)∫1/[(x/√2
+1/√2)²+1]d(x/√2+
1/√2)
=2ln|x-1|-½ln(x²+2x+3)+(√2/2)arctan[(x+1)/√2]
+C
