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

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(A) Given the partial fraction decomposition: $\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$ Multiplying b

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$A+C$

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(A) Given the partial fraction decomposition: $\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$ Multiplying both sides by $x^4(x+2)$, we get: $x+1 = Ax^3(x+2) + Bx^2(x+2) + Cx(x+2) + D(x+2) + Ex^4$ To find the value of $B+D+E$, we substitute $x = -1$ into the equation: $-1+1 = A(-1)^3(-1+2) + B(-1)^2(-1+2) + C(-1)(-1+2) + D(-1+2) + E(-1)^4$ $0 = A(-1)(1) + B(1)(1) + C(-1)(1) + D(1) + E(1)$ $0 = -A + B - C + D + E$ Rearranging the terms to isolate $B+D+E$: $B+D+E = A+C$

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

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