The partial fraction decomposition of frac{x^ | vedclass

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Question

(A) Let $u = x^2 + 1$,then $x^2 = u - 1$.Substituting this into the numerator: $x^4 + 24x^2 + 28 = (u - 1)^2 + 24(u - 1) + 28$.$= (u^2 - 2u + 1) + 24u

Vedclass · Accepted Answer

$\frac{1}{(x^2 + 1)} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$

Vedclass · Answer

(A) Let $u = x^2 + 1$,then $x^2 = u - 1$.Substituting this into the numerator: $x^4 + 24x^2 + 28 = (u - 1)^2 + 24(u - 1) + 28$.$= (u^2 - 2u + 1) + 24u - 24 + 28 = u^2 + 22u + 5$.Now,the expression becomes $\frac{u^2 + 22u + 5}{u^3}$.$= \frac{u^2}{u^3} + \frac{22u}{u^3} + \frac{5}{u^3} = \frac{1}{u} + \frac{22}{u^2} + \frac{5}{u^3}$.Substituting $u = x^2 + 1$ back,we get $\frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3}$.

The partial fraction decomposition of $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3}$ is:

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