If frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+frac{C | vedclass

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(D) Given the expression: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.First,perform polynomial division of $x

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(D) Given the expression: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.First,perform polynomial division of $x^4+3x+1$ by $(x+1)^2(x-1) = (x^2+2x+1)(x-1) = x^3+x^2-x-1$.Dividing $x^4+3x+1$ by $x^3+x^2-x-1$ gives a quotient of $(x-1)$ and a remainder of $3x^2+3x$. So,$Ax+B = x-1$,which implies $A=1$ and $B=-1$.Now,$\frac{3x^2+3x}{(x+1)^2(x-1)} = \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{x-1}$.Multiply by $(x+1)^2(x-1)$: $3x^2+3x = C(x+1)(x-1) + D(x-1) + E(x+1)^2$.For $x=1$: $3(1)^2+3(1) = E(1+1)^2 \Rightarrow 6 = 4E \Rightarrow E = 3/2$.For $x=-1$: $3(-1)^2+3(-1) = D(-1-1) \Rightarrow 0 = -2D \Rightarrow D = 0$.Comparing coefficients of $x^2$: $3 = C + E \Rightarrow 3 = C + 3/2 \Rightarrow C = 3/2$.Thus,$A=1, B=-1, C=3/2, D=0, E=3/2$.$A+B+C+D+E = 1 - 1 + 3/2 + 0 + 3/2 = 6/2 = 3$.

If $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$,then $A+B+C+D+E=$

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