frac{x^4}{(x^2+1)(x^2+3)} = | vedclass

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(D) To decompose the fraction $\frac{x^4}{(x^2+1)(x^2+3)}$,we first perform long division since the degree of the numerator is equal to the degree of

Vedclass · Accepted Answer

$1 + \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$ for some $A, B, C, D \in \mathbb{R}$

Vedclass · Answer

(D) To decompose the fraction $\frac{x^4}{(x^2+1)(x^2+3)}$,we first perform long division since the degree of the numerator is equal to the degree of the denominator.$\frac{x^4}{(x^2+1)(x^2+3)} = \frac{x^4}{x^4+4x^2+3} = \frac{(x^4+4x^2+3) - (4x^2+3)}{x^4+4x^2+3} = 1 - \frac{4x^2+3}{(x^2+1)(x^2+3)}$.Now,we use partial fractions for $\frac{4x^2+3}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$.Thus,the expression becomes $1 + \frac{-Ax-B}{x^2+1} + \frac{-Cx-D}{x^2+3}$,which is of the form $1 + \frac{Px+Q}{x^2+1} + \frac{Rx+S}{x^2+3}$.

$\frac{x^4}{(x^2+1)(x^2+3)} =$

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