{{{x^2} + 1} over {(2x - 1),({x^2} - 1)}} = | vedclass

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Question

(b) ${{{x^2} + 1} \over {(2x - 1)\,({x^2} - 1)}} = {A \over {(2x - 1)}} + {B \over {x + 1}} + {C \over {x - 1}}$

==> ${x^2} + 1 = A({x^2} - 1) +

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) ${{{x^2} + 1} \over {(2x - 1)\,({x^2} - 1)}} = {A \over {(2x - 1)}} + {B \over {x + 1}} + {C \over {x - 1}}$

==> ${x^2} + 1 = A({x^2} - 1) + B(2x - 1)\,(x - 1) + C(x + 1)\,(2x - 1)$

For $x = 1,$ $2 = 2C \Rightarrow C = 1$

For $x = - 1$, $2 = 6B \Rightarrow B = {1 \over 3}$

For $x = {1 \over 2}$, ${5 \over 4} = - {3 \over 4}A \Rightarrow $$A = - {5 \over 3}$

$	herefore $ Given expression = $ - {5 \over 3}{1 \over {(2x - 1)}} + {1 \over 3}{1 \over {x + 1}} + {1 \over {x - 1}}$

${{{x^2} + 1} \over {(2x - 1)\,({x^2} - 1)}} = $

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