જો ${x \over {(x - 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + y$ તો $y =$
Medium
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a
(a) \({x \over {(x - 1){{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + y\)
(a) \({x \over {(x - 1){{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + y\)
==>\({x \over {(x - 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + {{Ax + B} \over {{{({x^2} + 1)}^2}}}\)
==> \(4x = {({x^2} + 1)^2} - (x + 1)\,(x - 1)\,({x^2} + 1) + 4(Ax + B)\,(x - 1)\)
==> \(4A + 2 = 0\),\(4B - 4A = 4\) ==> \(A = {{ - 1} \over 2}\), \(B = {1 \over 2}\)
\(\therefore y = {{Ax + B} \over {{{({x^2} + 1)}^2}}} = {1 \over 2}{{(1 - x)} \over {{{({x^2} + 1)}^2}}}\)
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