If frac{3x^2 + 5}{(x^2 + 1)^2} = frac{a}{x^2 | vedclass

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(B) Given the partial fraction decomposition: $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$ Multiplying both sides by $(x

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$(3, 2)$

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(B) Given the partial fraction decomposition: $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$ Multiplying both sides by $(x^2 + 1)^2$,we get: $3x^2 + 5 = a(x^2 + 1) + b$ Expanding the right side: $3x^2 + 5 = ax^2 + a + b$ Comparing the coefficients of $x^2$ and the constant terms on both sides: $a = 3$ $a + b = 5$ Substituting $a = 3$ into the second equation: $3 + b = 5 \Rightarrow b = 2$ Therefore,$(a, b) = (3, 2)$.

If $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$,then $(a, b) = $

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