Question

2x+1/x^3+3x^2+2x resolve partial fraction​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

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$\sf \: \: \:   ⟼\dfrac{2x + 1}{ {x}^{3} + 3 {x}^{2} + 2x }$

$\sf \: \: \:   ⟼\dfrac{2x + 1}{x \: ( {x}^{2 } + 3x + 2) }$

$\sf \: \:   ⟼\dfrac{2x + 1}{x \: ( {x}^{2} + 2x + x + 2)}$

$\sf \: \:   ⟼\dfrac{2x + 1}{x \bigg(x(x + 2) + 1(x + 2) \bigg)}$

$\sf \:  \: ⟼\dfrac{2x + 1}{x \: (x + 2) \: (x + 1)}$

$\sf \:  ⟼Let \: \dfrac{2x + 1}{x \: (x + 1) \: (x + 2)} = \dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2} \sf \:  ⟼ \: (1)$

☆ On taking LCM as x(x+1)(x+2), we get

$\sf \: 2x + 1 = A(x + 1)(x + 2) + B(x)(x + 2) + C(x)(x + 1)\sf \:  ⟼(2)$

$\large\underline\red{\bold{❥︎Step :- 1 }}$

$\sf \: Substituting \: 'x = 0' \: in \: the \: equation \: (2), \: we \: get$

$\sf \:  ⟼1 = A(1)(2)$

$\bf\implies \:A = \dfrac{1}{2} \sf \:  ⟼ \: (3)$

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$\large\underline\red{\bold{❥︎Step :- 2 }}$

$\sf \: Substituting \: 'x = - 1' \: in \: the \: equation \: (2), \: we \: get$

$\sf \:  ⟼ - 2 \: + \: 1 = B \: ( - 1) \: (1)$

$\bf\implies \:B = 1 \: \sf \:  ⟼ \: (4)$

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

$\sf \:  Substituting \: 'x = - 2' \: in \: the \: equation \: (2), \: we \: get$

$\sf \:  ⟼ \: - 4 + 1 = C( - 2)( - 1)$

$\bf\implies \:C = - \dfrac{3}{2} \: \sf \:  ⟼ \: (5)$

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$\sf \: Substituting \: equation \: (3), \: (4), \: (5)\: in \: the \: equation \: (1), \: we \: get$

$\sf \:  \therefore \: \dfrac{2x + 1}{x \: (x + 1) \: (x + 2)} = \dfrac{1}{2 \: x} + \dfrac{1}{x + 1} - \dfrac{3}{2 \: (x + 2)} \sf \:$

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