(d) ${{{x^2} + 1} \over {({x^2} + 4)\,(x – 2)}} = {{Ax + B} \over {{x^2} + 4}} + {C \over {x – 2}}$

$ \Rightarrow $${x^2} + 1 = (Ax + B)\,(x – 2)\, + C({x^2} + 4)$$ \Rightarrow $$1 = A + C$

$ – 2A + B = 0$, $1 = – 2B + 4C$

$\therefore A = {3 \over 8},\,B = {3 \over 4},\,C = {5 \over 8}$

$\therefore {{{x^2} + 1} \over {({x^2} + 4)\,\,(x – 2)}} = {{{3 \over 8}x + {3 \over 4}} \over {{x^2} + 4}} + {{{5 \over 8}} \over {x – 2}}$

$ = \,{1 \over 4}\,\left( {{3 \over 8}x + {3 \over 4}} \right)\,{\left( {1 + {{{x^2}} \over 4}} \right)^{ – 1}} + {5 \over 8}\,{1 \over {( – 2)}}\,{\left( {1 – {x \over 2}} \right)^{ – 1}}$

$ = {1 \over 4}\,\left( {{3 \over 8}x + {3 \over 4}} \right)\,\,\left( {1 – {{{x^2}} \over 4} + {{\left( {{{{x^2}} \over 4}} \right)}^2} – {{\left( {{{{x^2}} \over 4}} \right)}^3} + ….} \right)$

$ – {5 \over {16}}\left( {1 + {x \over 2} + {{\left( {{x \over 2}} \right)}^2} + …..} \right)$

Coefficient of ${x^5}$= ${3 \over {32}}\,.\,{1 \over {{4^2}}} + {3 \over {16}} \times 0 – {5 \over {16}}\,{\left( {{1 \over 2}} \right)^5}$

= ${3 \over {{2^9}}} – {5 \over {{2^9}}} = – {1 \over {{2^8}}} = – {1 \over {256}}$ .