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

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Question

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

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$\frac{7}{4}$

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(C) Given the partial fraction decomposition: $\frac{3x + 4}{(x + 1)^2(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^2}$ Multiplying both sides by $(x + 1)^2(x - 1)$,we get: $3x + 4 = A(x + 1)^2 + B(x + 1)(x - 1) + C(x - 1)$ To find $A$,substitute $x = 1$ into the equation: $3(1) + 4 = A(1 + 1)^2 + B(1 + 1)(1 - 1) + C(1 - 1)$ $7 = A(2)^2 + 0 + 0$ $7 = 4A$ $A = \frac{7}{4}$

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

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