Question

Q. x² + 1/x(x²-1) resolve this in partial fraction
step by step solution​

Answer 1

Resolve in to partial fraction

$\bf \:\dfrac{ {x}^{2} + 1 }{x( {x}^{2} - 1)}$

Step by step explanation :-

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

$\: \bf \:\dfrac{ {x}^{2} + 1 }{x( {x}^{2} - 1)} = \bf \:\dfrac{ {x}^{2} + 1 }{x( {x} - 1)(x + 1)}$

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

☆ Taking x(x - 1)(x + 1) on both sides, we get

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

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

$\sf \:  ⟼On \: substituting \: x = 0, \: in \: equation \: (2), \: we \: get$

$\sf \:  ⟼1 = A( 0- 1)(0 + 1)$

$\bf\implies \:A = - 1 \: \bf \:  ⟼ (3)$

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

$\sf \:  ⟼On \: substituting \: x = 1, in \: equation \: (2), \: we \: get$

$\sf \:  ⟼ 1 + 1 = B(1)(1 + 1)$

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

$\large\underline{\bold{❥︎Step :- 4 }}$

$\sf \:  ⟼On \: substituting \: x = - 1, \: in \: equation \: (2), \: we \: get$

$\sf \:  ⟼ {( - 1)}^{2} + 1 = C( - 1)( - 1 - 1)$

$\sf \:  ⟼2 = 2C$

$\bf\implies \:C = 1 \: \bf \:  ⟼ \: (5)$

$\large\underline{\bold{❥︎Step :- 5 }}$

$\sf \:  ⟼On \: substituting \: A, B \: and \: C \: in \: equation (2), \: we \: get$

$\bf \ \: \bf \:\dfrac{ {x}^{2} + 1 }{x( {x} - 1)(x + 1)} = \dfrac{ - 1}{x} + \dfrac{1}{x - 1} + \dfrac{1}{x + 1}$

$\bf \therefore \: \bf \:\dfrac{ {x}^{2} + 1 }{x( {x}^{2} - 1)} = \dfrac{ - 1}{x} + \dfrac{1}{x - 1} + \dfrac{1}{x + 1}$

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