The partial fraction of x2+13x+15(2x+3) (x+3)2=...............

x2+13x+15(2x+3)(x+3)2=A2x+3 +Bx+3+C(x+3)2...............(i) ⇒A(x+3)2+B(2x+3)(x+3)+C(2x+3)=x2+13x+15If x=−3, C(−3)=9−39+15=−15 , C=5If x=−32, A(−32+3)2=94−392+15⇒9A4 =9−78+604=−94; A=−1 Coefficient of x2, A+2B=1 −1+2B=1, 2B=2, B=1(1) becomes −12x+3 +1x+3+5(x+3)2 =−1(2x+3) +1(x+3)+5(x+3)2

The partial fraction of x2+13x+15(2x+3) (x+3)2=...............

x2+13x+15(2x+3)(x+3)2=A2x+3 +Bx+3+C(x+3)2...............(i) ⇒A(x+3)2+B(2x+3)(x+3)+C(2x+3)=x2+13x+15If x=−3, C(−3)=9−39+15=−15 , C=5If x=−32, A(−32+3)2=94−392+15⇒9A4 =9−78+604=−94; A=−1 Coefficient of x2, A+2B=1 −1+2B=1, 2B=2, B=1(1) becomes −12x+3 +1x+3+5(x+3)2 =−1(2x+3) +1(x+3)+5(x+3)2

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The partial fraction of $\frac{x^{2} + 13 x + 15}{(2 x + 3) {(x + 3)}^{2}} = ...............$

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

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

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

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

answer is B.

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Detailed Solution

$\frac{x^{2} + 13 x + 15}{(2 x + 3) {(x + 3)}^{2}} = \frac{A}{2 x + 3} + \frac{B}{x + 3} + \frac{C}{{(x + 3)}^{2}} ............... (i) \begin{array}{l} \Rightarrow A {(x + 3)}^{2} + B (2 x + 3) (x + 3) + C (2 x + 3) = x^{2} + 13 x + 15 \\ I f x = - 3, C (- 3) = 9 - 39 + 15 = - 15, C = 5 \\ I f x = \frac{- 3}{2}, A {(\frac{- 3}{2} + 3)}^{2} = \frac{9}{4} - \frac{39}{2} + 15 \\ \Rightarrow \frac{9 A}{4} = \frac{9 - 78 + 60}{4} = \frac{- 9}{4}; A = - 1 \end{array}$

Coefficient of $x^{2}, A + 2 B = 1$

$\begin{array}{l} - 1 + 2 B = 1, 2 B = 2, B = 1 \\ (1) b e c o m e s \frac{- 1}{2 x + 3} + \frac{1}{x + 3} + \frac{5}{{(x + 3)}^{2}} = \frac{- 1}{(2 x + 3)} + \frac{1}{(x + 3)} + \frac{5}{{(x + 3)}^{2}} \end{array}$