If ${{ax + b} \over {{{(3x + 4)}^2}}} = {1 \over {3x + 4}} - {3 \over {{{(3x + 4)}^2}}}$ then
$a = 2$
$b = 1$
$a = 3$
$(b)$ and $(c)$ both
(d) $ax + b = (3x + 4) – 3$ $ \Rightarrow $ $a = 3,\,b = 4 – 3 = 1$.
The partial fractions of ${{3x – 1} \over {(1 – x + {x^2})\,(2 + x)}}$ are
If ${{3x + a} \over {{x^2} – 3x + 2}} = {A \over {(x – 2)}} – {{10} \over {x – 1}}$, then
${{(x – a)(x – b)} \over {(x – c)(x – d)}} = {A \over {x – c}} – {B \over {(x – d)}} + C$, then $C =$
${1 \over {x({x^2} + 1)}} = {A \over x} + {{Bx + C} \over {({x^2} + 1)}}$, then $(A,\,B,\,C) = $
${{3{x^2} + 5} \over {{{({x^2} + 1)}^2}}} = {a \over {{x^2} + 1}} + {b \over {{{({x^2} + 1)}^2}}}$, then $(a,b) = $
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