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I still have not grasped the intuitive reason for adding a coefficient for each degree in denominator when doing partial fraction integration. For example

$\int_a^b \frac {R(x)}{(x+2)^3} $ , we need to do $\frac {A}{(x+2)^1} + \frac {B}{(x+2)^2} + \frac {C}{(x+2)^3}$.

Why?

Haim
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  • See also here: http://math.stackexchange.com/questions/672430/partial-fraction-decomposition, http://math.stackexchange.com/questions/665712/extra-square-in-partial-fraction – Hans Lundmark May 29 '16 at 20:16

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This results from Taylor's formula, which is an exact formula for polynomials.

Remember $R(x)$ has degree $\le 3$, hence we have $$R(x)=R(-2)+R'(-2)(x+2)+r''(-2)\frac{(x+2)^2}2+r'''(-2)\frac{(x+2)^3}6.$$ You just have to divide both sides by $(x+2)^3$.

Bernard
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