Which of the following is a partial fraction | vedclass

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(B) Given expression is $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} = \frac{-x^{2} + 6x + 13}{(3x + 5)(x + 2)^{2}}$.Using partial fraction deco

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

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(B) Given expression is $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} = \frac{-x^{2} + 6x + 13}{(3x + 5)(x + 2)^{2}}$.Using partial fraction decomposition,we write:$\frac{-x^{2} + 6x + 13}{(3x + 5)(x + 2)^{2}} = \frac{A}{3x + 5} + \frac{B}{x + 2} + \frac{C}{(x + 2)^{2}}$.Multiplying both sides by $(3x + 5)(x + 2)^{2}$,we get:$-x^{2} + 6x + 13 = A(x + 2)^{2} + B(x + 2)(3x + 5) + C(3x + 5)$.To find $C$,put $x = -2$:$-(-2)^{2} + 6(-2) + 13 = C(3(-2) + 5) \Rightarrow -4 - 12 + 13 = C(-1) \Rightarrow -3 = -C \Rightarrow C = 3$.To find $A$,put $x = -\frac{5}{3}$:$-(-\frac{5}{3})^{2} + 6(-\frac{5}{3}) + 13 = A(-\frac{5}{3} + 2)^{2} \Rightarrow -\frac{25}{9} - 10 + 13 = A(\frac{1}{3})^{2} \Rightarrow \frac{2}{9} = A(\frac{1}{9}) \Rightarrow A = 2$.To find $B$,compare the coefficients of $x^{2}$ on both sides:$-1 = A + 3B \Rightarrow -1 = 2 + 3B \Rightarrow -3 = 3B \Rightarrow B = -1$.Thus,the partial fraction is $\frac{2}{3x + 5} - \frac{1}{x + 2} + \frac{3}{(x + 2)^{2}}$.

Which of the following is a partial fraction of $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} =$

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