定积分题一道,

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定积分题一道,
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定积分题一道,
定积分题一道,

定积分题一道,

使用奥斯特洛格拉德斯基的待定系数法
设1/((x^2+1)(x^2+4)^2)= (Ax+B)/(x^2+1)+(Cx+D)/(x^2+4)+(Ex+F)/(x^2+4)^2
则:(Ax+B)(x^2+4)^2+(Cx+D)(x^2+1)(x^2+4)+(Ex+F)(x^2+1)=1
(Ax+B)(x^4+8x^2+16)+(Cx+D)(x^4+5x^2+4)+(E...

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使用奥斯特洛格拉德斯基的待定系数法
设1/((x^2+1)(x^2+4)^2)= (Ax+B)/(x^2+1)+(Cx+D)/(x^2+4)+(Ex+F)/(x^2+4)^2
则:(Ax+B)(x^2+4)^2+(Cx+D)(x^2+1)(x^2+4)+(Ex+F)(x^2+1)=1
(Ax+B)(x^4+8x^2+16)+(Cx+D)(x^4+5x^2+4)+(Ex+F)(x^2+1)=1
Ax^5+8Ax^3+16Ax+Bx^4+8Bx^2+16B +Cx^5+5Cx^3+4Cx+ Dx^4+5Dx^2+4D+Ex^3+Ex+Fx^2+F=1
(A+C)x^5+(B+D)x^4+(8A+5C+E)x^3+(8B+5D+F)x^2+(16A+4C+E)x+16B+4D+F=1
A+C=0 B+D=0 8A+5C+E=0 8B+5D+F=0 16A+4C+E=0 16B+4D+F=1
联立解得:A=0 B=1/9 C=0 D=-1/9 E=0 F=-1/3
原式=1/9∫[-∞, +∞][1/(x^2+1)-1/(x^2+4)-3/(x^2+4)^2]dx
=1/9[arctanx-1/2 arctan(x/2)-3/8*x/(x^2+4)-3/16*atan(1/2*x)][-∞, +∞]
=1/9[π-π/2-3π/16]
=5π/134
计算量太大,您再核实一下数据。

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