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

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

Vedclass · Accepted Answer

$1.75$

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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 the denominator ${(x + 1)}^2(x - 1)$,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} = 1.75$

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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