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