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Resolve `(2x+1)/((x+3)(x^(2)+1)^(2))` into partial fractions. |
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Answer» Let, `(2x+1)/((x+3)(x^(2)+1)^(2))=(A)/(x+3)+(Bx+C)/(x^(2)+1)+(Dx+E)/((x^(2)+1)^(2))` `implies2x+1=A(x^(2)+1)^(2)+(Bx+C)(x+3)(x^(2)+1)+(Dx+E)(x+3)`………`(1)` Put `x=-3` in Eq. `(1)`, we have `-5=100AimpliesA=(-1)/(20)` Comparing the coefficients of `x^(4)` on either sides of Eq. `(1)`, we have `0=A+BimpliesB=(1)/(20)` Compairng the coefficients of `x^(3)` on either sides of Eq. `(1)`, we have `0=3B+CimpliesC=(-3)/(20)` Put `x=0` in Eq. `(1)`, we have `implies1=A+3C+3Eimplies3E=1+(1)/(20)+(9)/(20)` `impliesE=(1)/(2)` By putting `x=1` in Eq. `(1)`, we have `3=4A+8B+8C+4D+4E` `3=(-4)/(20)+(8)/(20)+8((-3)/(20))+4D+((1)/(2))` `3+(4)/(20)-(8)/(20)+(24)/(20)-2=4D` `D=(2)/(4)=(1)/(2)` `:.(2x+1)/((x+3)(x^(2)+1)^(2))=((-1)/(20))/(x+3)+((x)/(20)-(3)/(20))/(x^(2)+1)+((1)/(2)x+(1)/(2))/((x^(2)+1)^(2))` `=(-1)/(20(x+3))+((x-3))/(20(x^(2)+1))+((x+1))/(2(x^(2)+1)^(2))` |
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