${{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = $
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a
(a) \({{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = {A \over {2x + 3}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}\)
(a) \({{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = {A \over {2x + 3}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}\)
==> \({x^2} + 13x + 15 = A{(x + 3)^2} + B(2x + 3)\,(x + 3) + C(2x + 3)\)
For \(x = - 3,\,C = 5\) and for \(x = - {3 \over 2};\,A = - 1\)
Equating coefficient of \({x^2}\)
\(1 = A + 2B \Rightarrow B = {{1 - A} \over 2} = 1\)
\(\therefore \) Given expression = \({1 \over {x + 3}} - {1 \over {2x + 3}} + {5 \over {{{(x + 3)}^2}}}\).
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